Question: As shown in the figure, the line $y=-\frac{1}{2}x+2$ intersects the x-axis at point B and the y-axis at point C. A parabola with the axis of symmetry $x=\frac{3}{2}$ passes through points B and C, and intersects the negative half of the x-axis at point A. What is the analytic equation of the parabola?

Solution:
\textbf{Solution:} Let $x=0$, then $y=2$. Let $y=0$, then $-\frac{1}{2}x+2=0$, which implies $x=4$,
\[\therefore B(4, 0), C(0, 2),\]
\[\because\] the axis of symmetry of the parabola is $x=\frac{3}{2}$,
\[\therefore\] the parabola can be expressed as $y=a\left(x-\frac{3}{2}\right)^{2}+k$,
\[\because\] the parabola passes through points $B(4, 0)$, $C(0, 2)$,
\[\therefore \left\{\begin{array}{l}\frac{25}{4}a+k=0 \\ \frac{9}{4}a+k=2\end{array}\right.,\]
Solving this, we get \[\left\{\begin{array}{l}a=-\frac{1}{2} \\ k=\frac{25}{8}\end{array}\right.,\]
\[\therefore\] the equation of the parabola is \[y=-\frac{1}{2}\left(x-\frac{3}{2}\right)^{2}+\frac{25}{8}=-\frac{1}{2}x^{2}+\frac{3}{2}x+2=\boxed{-\frac{1}{2}x^{2}+\frac{3}{2}x+2}.\]


Question: As shown in the diagram, the line $y=-\frac{1}{2}x+2$ intersects the x-axis at point B and the y-axis at point C. A parabola with the axis of symmetry $x=\frac{3}{2}$ passes through points B and C and intersects the negative x-axis at point A. Let P be a point on the parabola. Connect AP, and rotate line segment AP clockwise by $90^\circ$ around point A to obtain line segment AQ. When the distance from point Q to the axis of symmetry is $\frac{1}{2}$, what are the coordinates of point P?

Solution:
\textbf{Solution:}

Let $PH \perp AB$, $QG \perp AB$, and $QM \perp l$, where H, G, M are the feet of the perpendiculars. Due to the symmetry of the parabola, it can be derived that $A(-1, 0)$,

since $AP = AQ$, $\angle APH = \angle QAG = 90^\circ - \angle PAH$, $\angle AHP = \angle QGA = 90^\circ$,

thus $\triangle AHP \cong \triangle QGA \ (AAS)$,

therefore, $PH = AG$,

since the distance from A to the axis of symmetry is $\frac{5}{2}$ and the distance from Q to the axis of symmetry is $\frac{1}{2}$,

when Q is on the left side of the axis of symmetry, $PH = AG = 2$,

when Q is on the right side of the axis of symmetry, $PH = AG = 3$,

therefore, $-\frac{1}{2}x^2 + \frac{3}{2}x + 2 = 2$ or $-\frac{1}{2}x^2 + \frac{3}{2}x + 2 = 3$,

solving yields $x_1 = 0$, $x_2 = 3$ or $x_1 = 1$, $x_1 = 2$

thus, the coordinates of point P are $\boxed{(0, 2)}$ or $\boxed{(3, 2)}$ or $\boxed{(1, 3)}$ or $\boxed{(2, 3)}$;


Question: As shown in the diagram, the line $y = -\frac{1}{2}x + 2$ intersects the x-axis at point B and the y-axis at point C. A parabola with the axis of symmetry $x = \frac{3}{2}$ passes through points B and C and intersects the negative half of the x-axis at point A. Let M be a moving point on the parabola, and N be on the line $BC$. When the quadrilateral formed by points O, C, M, and N is a parallelogram, directly write down the coordinates of point N.

Solution:
\textbf{Solution:} Let $N_1(2, 1)$, $N_2\left(2+2\sqrt{2}, 1-\sqrt{2}\right)$, $N_3\left(2-2\sqrt{2}, 1+\sqrt{2}\right)$, $N_4\left(-2\sqrt{2}-2, 3+\sqrt{2}\right)$, and $N_5\left(-2+2\sqrt{2}, 3-\sqrt{2}\right)$.

\boxed{N_1(2, 1)}, \boxed{N_2\left(2+2\sqrt{2}, 1-\sqrt{2}\right)}, \boxed{N_3\left(2-2\sqrt{2}, 1+\sqrt{2}\right)}, \boxed{N_4\left(-2\sqrt{2}-2, 3+\sqrt{2}\right)}, and \boxed{N_5\left(-2+2\sqrt{2}, 3-\sqrt{2}\right)}.


Question: As shown in the figure, the graph of the quadratic function $y=ax^2+bx+c$ intersects the x-axis at points B and C, and intersects the y-axis at point A. If the coordinates of point A are $(0, -3)$, $\angle ABC=45^\circ$, and $3OC=OA$, find the expression for this quadratic function.

Solution:
\textbf{Solution:} Given the coordinates of point A, we know that $AO=3$, \\
$\because$ $\angle ABC=45^\circ$ and $OC=OA$, \\
$\therefore$ $BO=3$, $CO=1$, \\
Therefore, the coordinates of points B and C are $(-3,0)$ and $(1,0)$, respectively, \\
Let the equation of the parabola be $y=a(x-x_1)(x-x_2)=a(x-1)(x+3)$, \\
$\because$ the coordinates of point A are $(0,-3)$, thus $-3a=-3$, solving this gives $a=1$, \\
Therefore, the equation of the parabola is $y=x^2+2x-3=\boxed{x^2+2x-3}$.


Question: As shown in the figure, it is the graph of the quadratic function $y=ax^2+bx+c$, which intersects the x-axis at points B and C, and intersects the y-axis at point A. Point P is on the parabola, and the area of $\triangle BCP$ is $\frac{4}{3}$ times the area of $\triangle ABC$. What are the coordinates of point P?

Solution:
\textbf{Solution:} Because $S_{\triangle BCP} = \frac{4}{3}S_{\triangle ABC}$, \\
it follows that $|y_P| = \frac{4}{3}|y_C|$, which means $y_P = \pm 4$, \\
i.e. $y = x^2 + 2x - 3 = \pm 4$, \\
Solving this gives: $x= -1 \pm 2\sqrt{2}$ or $-1$, \\
Hence, the coordinates of point P are $\boxed{(-1+2\sqrt{2},4)}$ or $\boxed{(-1-2\sqrt{2},4)}$ or $\boxed{(-1,-4)}$.


Question: As shown in the Cartesian coordinate system $xOy$, A and B are two points on the x-axis, while C and D are two points on the y-axis. The parabolic segment $C_1$ passing through points A, C, B and the parabolic segment $C_2$ passing through points A, D, B combine to form a closed curve, which we call the “egg curve”. It is known that the coordinates of point C are $(0,)$, and point M is the vertex of the parabola $C_2\colon y=mx^2-2mx-3m$ ($m<0$). What are the coordinates of points A and B?

Solution:
\textbf{Solution:} Let $y=0$, then $mx^{2}-2mx-3m=0$,\\
since $m<0$, it follows that $x^{2}-2x-3=0$, solving this we get: $x_{1}=-1$, $x_{2}=3$.\\
Therefore, the coordinates of A are $\boxed{(-1,0)}$, and the coordinates of B are $\boxed{(3,0)}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, A and B are two points on the x-axis, C and D are two points on the y-axis. A segment of a parabola passing through points A, C, and B, denoted as $C_1$, and a segment of a parabola passing through points A, D, and B, denoted as $C_2$, combine to form a closed curve, which we call the "egg curve". It is known that the coordinate of point C is (0, ), and point M is the vertex of the parabola $C_2$: $y=mx^2-2mx-3m$ ($m<0$). Is there a point P on the "egg curve" in the fourth quadrant such that the area of $\triangle PBC$ is maximized? If such a point exists, find the maximum area of $\triangle PBC$; if not, provide a reason.

Solution:
\textbf{Solution:} There exists a solution. The reason is as follows:\\
$\because$ Let the expression of parabola $C_1$ be $y=\frac{1}{2}(x+1)(x-3)$. Substituting $C(0, -\frac{3}{2})$ into it, we get $a=\frac{1}{2}$.\\
$\therefore$ The expression of $C_1$ is: $y=\frac{1}{2}x^2-x-\frac{3}{2}$.\\
Let $P(p, \frac{1}{2}p^2-p-\frac{3}{2})$,\\
$\therefore S_{\triangle PBC}=S_{\triangle POC}+S_{\triangle BOP}-S_{\triangle BOC}=-\frac{3}{4}\left(p-\frac{3}{2}\right)^2+\frac{27}{16}$.\\
$\because a=-\frac{3}{4}<0$, $\therefore$ when $p=\frac{3}{2}$, the maximum value of $S_{\triangle PBC}$ is $\boxed{\frac{27}{16}}$.


Question: In the Cartesian coordinate system $xOy$, A and B are two points on the x-axis, and C and D are two points on the y-axis. A part of the parabola passing through points A, C, and B, denoted as $C_{1}$, and a part of the parabola passing through points A, D, and B, denoted as $C_{2}$, combine to form a closed curve. We call this closed curve an "egg curve". Given that the coordinates of point C are (0,), and point M is the vertex of the parabola $C_{2}\colon y=mx^{2}-2mx-3m$ ($m<0$). When $\triangle BDM$ is a right-angled triangle, what is the value of $m$?

Solution:
\textbf{Solution:} From $C_{2}$, it is known that: $B(3,0)$, $D(0,-3m)$, $M(1,-4m)$, \\
$\therefore BD^{2}=9m^{2}+9$, $BM^{2}=16m^{2}+4$, $DM^{2}=m^{2}+1$. \\
$\because$ $\angle MBD<90^\circ$, $\therefore$ we discuss the cases of $\angle BMD=90^\circ$ and $\angle BDM=90^\circ$: \\
When $\angle BMD=90^\circ$, $BM^{2}+DM^{2}=BD^{2}$, i.e., $16m^{2}+4+m^{2}+1=9m^{2}+9$, \\
we obtain: $m_{1}=-\frac{\sqrt{2}}{2}$, $m_{2}=\frac{\sqrt{2}}{2}$ (discard). \\
When $\angle BDM=90^\circ$, $BD^{2}+DM^{2}=BM^{2}$, i.e., $9m^{2}+9+m^{2}+1=16m^{2}+4$, \\
we obtain: $m_{1}=-1$, $m_{2}=1$ (discard). \\
In summary, when $m=-\frac{\sqrt{2}}{2}$ or $m=-1$, $\triangle BDM$ is a right-angled triangle. \\
Therefore, the value of $m$ is $\boxed{-\frac{\sqrt{2}}{2}}$ or $\boxed{-1}$.


Question: As shown in the graph, the parabola $y=2x^{2}-4x-6$ intersects the x-axis at points A and B, with point A being to the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. Find the coordinates of points A, B, and C.

Solution:
\textbf{Solution:} Substitute $y=0$ into $y=2x^{2}-4x-6$,\\
we get $2x^{2}-4x-6=0$.\\
Solving it, we find $x_{1}=-1$, $x_{2}=3$.\\
Therefore, the coordinates of point A are $\boxed{\left(-1, 0\right)}$, and the coordinates of point B are $\boxed{\left(3, 0\right)}$.\\
Substitute $x=0$ into $y=2x^{2}-4x-6$, we get $y=-6$.\\
Therefore, the coordinates of point C are $\boxed{\left(0, -6\right)}$.


Question:
Synthesis and Practice\\
As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points $A$ and $B$, where point $A$ is to the left of point $B$, and intersects the y-axis at point $C$. Point $D$ is a moving point on the parabola. As shown in Figure 2, when point $D$ is in the fourth quadrant, connect $BD$, $CD$, and $BC$ to form $\triangle BCD$. When the area of $\triangle BCD$ is maximized, what are the coordinates of point $D$?

Solution:
Solution: Let the coordinates of point D be $\left(m, 2m^{2}-4m-6\right)$.

As shown in the diagram, draw $DH \perp x$-axis at point H, and draw $DG \perp y$-axis at point G, then connect $OD$.

$\therefore DG=m$, $DH=-2m^{2}+4m+6$.

$\because$ The coordinates of point B are $\left(3, 0\right)$, and the coordinates of point C are $\left(0, -6\right)$.

$\therefore OB=3$, $OC=6$.

$\because S_{\triangle BCD}=S_{\triangle OCD}+S_{\triangle OBD}-S_{\triangle OBC}$,

$\therefore S_{\triangle BCD}=\frac{OC\cdot DG}{2}+\frac{OB\cdot DH}{2}-\frac{OC\cdot OB}{2}$

$S_{\triangle BCD}=\frac{6m}{2}+\frac{3(-2m^{2}+4m+6)}{2}-\frac{6\times 3}{2}$

Simplifying, we get $S_{\triangle BCD}=-3\left(m-\frac{3}{2}\right)^{2}+\frac{27}{4}$.

$\because -3<0$,

$\therefore$ When $m=\frac{3}{2}$, the maximum area of $\triangle BCD$ is $\frac{27}{4}$.

$\therefore 2m^{2}-4m-6=2\times \left(\frac{3}{2}\right)^{2}-4\times \frac{3}{2}-6=2\times \frac{9}{4}-6-6=-\frac{15}{2}$.

$\therefore$ The coordinates of point D are $\boxed{\left(\frac{3}{2}, -\frac{15}{2}\right)}$.


Question: As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points A and B, with point A to the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. Find the coordinates of points A, B, and C.

Solution:
\textbf{Solution:} Substitute $y=0$ into $y=2x^{2}-4x-6$,
we have $2x^{2}-4x-6=0$.\\
Solving this, we find $x_{1}=-1$ and $x_{2}=3$.\\
Therefore, the coordinates of point A are $\boxed{\left(-1, 0\right)}$, and the coordinates of point B are $\boxed{\left(3, 0\right)}$.\\
Substituting $x=0$ into $y=2x^{2}-4x-6$, we get $y=-6$.\\
Therefore, the coordinates of point C are $\boxed{\left(0, -6\right)}$.


Question:
Comprehensive and Practical

As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points A and B, with point A on the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. As shown in Figure 2, when point D is in the fourth quadrant, connect $BD$, $CD$, and $BC$ to form $\triangle BCD$. What are the coordinates of point D when the area of $\triangle BCD$ is maximized?

Solution:
\textbf{Solution:} Let the coordinates of point D be $\left(\frac{3}{2}, 2m^2-4m-6\right)$.

Draw $DH \perp x$-axis at point H, and draw $DG \perp y$-axis at point G, connecting $OD$.

$\therefore DG=m$, $DH=-2m^2+4m+6$.

$\because$ The coordinates of point B are $(3, 0)$ and the coordinates of point C are $(0, -6)$.

$\therefore OB=3$, $OC=6$.

$\because S_{\triangle BCD}=S_{\triangle OCD}+S_{\triangle OBD}-S_{\triangle OBC}$,

$\therefore S_{\triangle BCD}=\frac{OC \cdot DG}{2}+\frac{OB \cdot DH}{2}-\frac{OC \cdot OB}{2}$

$S_{\triangle BCD}=\frac{6m}{2}+\frac{3(-2m^2+4m+6)}{2}-\frac{6 \times 3}{2}$

Simplifying, we get $S_{\triangle BCD}=-3\left(m-\frac{3}{2}\right)^2+\frac{27}{4}$.

$\because -3<0$,

$\therefore$ When $m=\frac{3}{2}$, the area of $\triangle BCD$ is maximized at $\frac{27}{4}$.

$\therefore 2m^2-4m-6=2\times \left(\frac{3}{2}\right)^2-4\times \frac{3}{2}-6=2\times \frac{9}{4}-6-6=-\frac{15}{2}$.

$\therefore$ The coordinates of point D are $\boxed{\left(\frac{3}{2}, -\frac{15}{2}\right)}$.


Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. What are the equation of the parabola and the coordinates of its vertex?

Solution:
\textbf{Solution:} Substituting the points $A(-1,0)$ and $B(3,0)$ into $y=x^{2}+bx+c$, \\
we get: $\begin{cases}1-b+c=0\\ 9+3b+c=0\end{cases}$, \\
Solving, we find: $\begin{cases}b=-2\\ c=-3\end{cases}$, \\
$\therefore$ The equation of the parabola is $y=x^{2}-2x-3$, \\
$\because y=x^{2}-2x-3=(x-1)^{2}-4$, \\
$\therefore$ The vertex coordinates are $\boxed{(1, -4)}$.


Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. What is the range of $y$ when $0<x<3$?

Solution:
\textbf{Solution:} From the graph, we can determine that when $0<x<3$,\\
we have $-4\le y<0$.\\
Therefore, the range of $y$ is $y\in\boxed{[-4,0)}$.


Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. Point $P$ is a point on the parabola. If the area of $\triangle PAB$ is $10$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Given $A(-1,0)$ and $B(3,0)$,\\
thus $AB=4$.\\
Let $P(x,y)$, then the area $S_{\triangle PAB}=\frac{1}{2}AB|y|=10$,\\
thus $|y|=5$,\\
therefore $y=\pm 5$.\\
(1) When $y=5$,\\
$x^{2}-2x-3=5$,\\
solving this, we find: $x_{1}=-2$, $x_{2}=4$,\\
thus, the coordinates of point $P$ are $(-2,5)$ or $(4, 5)$;\\
(2) When $y=-5$,\\
$x^{2}-2x-3=-5$,\\
this equation has no solution;\\
In conclusion, the coordinates of point $P$ are $\boxed{(-2,5)}$ or $\boxed{(4, 5)}$.


Question: As shown in the figure, the parabola $y=ax^{2}+2ax+1$ has only one common point A with the x-axis. The line passing through point A intersects the parabola at point B and the y-axis at point C, and point C is the midpoint of line segment AB. What is the analytical expression of the function corresponding to this parabola?

Solution:
\textbf{Solution:} Since the parabola $y=ax^{2}+2ax+1$ has only one common point $A$ with the x-axis,\\
therefore, $\Delta=4a^{2}-4a=0$, solving this yields $a_{1}=0$ (discard), $a_{2}=1$,\\
therefore, the equation of the parabola is $\boxed{y=x^{2}+2x+1}$.


Question: As shown in the figure, the parabola $y=ax^{2}+2ax+1$ has only one common point A with the x-axis. A line passing through point A intersects the parabola at point B and intersects the y-axis at point C, and point C is the midpoint of line segment AB. Find the analytical expression of the line AB.

Solution:
\textbf{Solution:}
Given that $y=(x+1)^{2}$,\\
hence, the coordinates of vertex A are (-1, 0),\\
since point C is the midpoint of segment AB,\\
meaning that points A and B are symmetric with respect to point C,\\
thus, the x-coordinate of point B is 1,\\
when $x=1$, $y=x^{2}+2x+1=1+2+1=4$, so point B is (1, 4),\\
let the equation of line AB be $y=kx+b$,\\
substituting A (-1, 0) and B (1, 4) into it, we get\\
$\begin{cases}
-k+b=0\\
k+b=4
\end{cases}$\\
Solving this, we find $\begin{cases}
k=2\\
b=2
\end{cases}$\\
thus, the equation of line AB is $\boxed{y=2x+2}$.


Question: As shown in the figure, the parabola $y=ax^2-3ax-4a$ passes through point $C(0, 2)$, intersects the $x$ axis at points $A$ and $B$ (with point $A$ to the left of point $B$), connects $BC$, and the line $y=kx+1$ ($k>0$) intersects the $y$ axis at point $D$, intersects the parabola above $BC$ at point $E$, and intersects $BC$ at point $F$. Find the analytical expression of the parabola and the coordinates of points $A$ and $B$.

Solution:
\textbf{Solution:}

Substitute $C\left(0, 2\right)$ into $y=a{x}^{2}-3ax-4a$, which gives $-4a=2$. Solving this, we find $a=-\frac{1}{2}$.

$\therefore$ The equation of the parabola is $y=-\frac{1}{2}{x}^{2}+\frac{3}{2}x+2$.

Setting $-\frac{1}{2}{x}^{2}+\frac{3}{2}x+2=0$,

we find:${x}_{1}=-1, {x}_{2}=4$.

$\therefore \boxed{A\left(-1, 0\right)}, \boxed{B\left(4, 0\right)}$


Question: As shown in the figure, the parabola $y=ax^2-3ax-4a$ passes through point $C(0, 2)$ and intersects the x-axis at points $A$ and $B$ (point $A$ is to the left of point $B$). Line $BC$ is drawn, and the line $y=kx+1$ ($k>0$) intersects the y-axis at point $D$ and intersects the parabola above line $BC$ at point $E$, and intersect line $BC$ at point $F$. Does $\frac{EF}{DF}$ have a maximum value? If it exists, please find the maximum value and the coordinates of point $E$ at that time; if it does not exist, please explain why.

Solution:
\textbf{Solution:} Existence is shown as follows,\\
As illustrated, according to the problem, point $E$ is on the right side of the $y$-axis. Construct $EG\parallel y$-axis, intersecting $BC$ at point $G$\\
$\therefore CD\parallel EG$\\
$\therefore \frac{EF}{DF}=\frac{EG}{CD}$\\
$\because$ The line $y=kx+1\,(k>0)$ intersects the $y$-axis at point $D$\\
$\therefore D(0, 1),$\\
$\therefore CD=2-1=1,$\\
$\therefore \frac{EF}{DF}=EG$\\
Assume the equation of the line where $BC$ lies is $y=mx+n\,(m\ne 0)$,\\
Substituting $B(4, 0), C(0, 2)$ into the above equation yields: $\left\{\begin{array}{l} 0=4m+n \\ 2=n \end{array}\right.$\\
Solving gives: $\left\{\begin{array}{l} m=-\frac{1}{2} \\ n=2 \end{array}\right.$\\
$\therefore$ The equation of line $BC$ is $y=-\frac{1}{2}x+2$\\
Let $E(t, -\frac{1}{2}t^2+\frac{3}{2}t+2)$\\
Then $G(t, -\frac{1}{2}t+2)$, where $0<t<4$.\\
$\therefore EG=-\frac{1}{2}t^2+\frac{3}{2}t+2-\left(-\frac{1}{2}t+2\right)=-\frac{1}{2}(t-2)^2+2$\\
$\therefore \frac{EF}{DF}=-\frac{1}{2}(t-2)^2+2,$\\
$\because -\frac{1}{2}<0,$\\
$\therefore$ The parabola opens downwards\\
$\therefore$ At $t=2$, it has the maximum value, which is $\boxed{2}$.\\
Substituting $t=2$ into $-\frac{1}{2}t^2+\frac{3}{2}t+2=-2+3+2=3$\\
$\therefore$ The coordinates of point $E$ are $\boxed{(2, 3)}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ ($a\ne 0$) has its vertex at $Q(2, -1)$, and intersects the $y$-axis at point $C(0, 3)$, and intersects the $x$-axis at points $A$ and $B$ (point $A$ is to the right of point $B$). Point $P$ is a moving point on the parabola, moving from point $C$ towards point $A$ along the parabola (point $P$ does not coincide with $A$). A perpendicular line is drawn from point $P$ to line $AC$ at point $D$. What is the equation of this parabola, and the coordinates of points $A$ and $B$?

Solution:
\textbf{Solution:} Given that the vertex of the parabola is $Q(2, -1)$,\\
thus, the equation of the parabola can be assumed as $y=a(x-2)^2-1$.\\
Substituting $C(0, 3)$ into $y=a(x-2)^2-1$,\\
we get: $3=a(0-2)^2-1$,\\
from which it follows that $a=\boxed{1}$.\\
Therefore, the equation of the parabola is $y=(x-2)^2-1=x^2-4x+3$,\\
Setting $y=0$, gives $x^2-4x+3=0$,\\
from which we find: $x_1=1$, $x_2=3$,\\
since point $A$ is to the right of point $B$,\\
therefore, $\boxed{A(3, 0)}$, $\boxed{B(1, 0)}$;


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ (where $a\ne 0$) has its vertex at the point $Q(2, -1)$, and it intersects the $y$-axis at point $C(0, 3)$, and intersects the $x$-axis at points $A$ and $B$ (with point $A$ being to the right of point $B$). Point $P$ is a moving point on the parabola, moving from point $C$ towards point $A$ (point $P$ does not coincide with $A$). A perpendicular line is drawn from point $P$ to line $AC$ at point $D$. What is the maximum value of segment $PD$ during the motion of point $P$?

Solution:
\textbf{Solution:} Let the equation of the line $AC$ be $y=kx+b$,\\
Substituting $A\left(3, 0\right)$ and $C\left(0, 3\right)$ into $y=kx+b$, we get:\\
$\begin{cases}
3k+b=0 \\
b=3
\end{cases}$, solving this we find $\begin{cases}
k=-1 \\
b=3
\end{cases}$,\\
$\therefore$ the equation of the line $AC$ is $y=-x+3$,\\
Draw $EF\perp x$ axis from point P, intersecting line $AC$ at point E and the $x$ axis at point F, connect $AP$ and $CP$,\\
$\because$ Point P is on the parabola, point E is on the line $AC$, and $EF\perp x$ axis,\\
Let $P\left(x, x^{2}-4x+3\right)$ and $E\left(x, -x+3\right)$,\\
$\therefore PE=-x+3-\left(x^{2}-4x+3\right)=-x^{2}+3x$,\\
$\therefore Area_{\triangle PAC}=Area_{\triangle PEC}+Area_{\triangle PEA}$,\\
$=\frac{1}{2}PE\times x+\frac{1}{2}PE\times (3-x)$,\\
$=\frac{3}{2}PE$,\\
$=-\frac{3}{2}x^{2}+\frac{9}{2}x$,\\
$\because a=-\frac{3}{2}<0$,\\
$\therefore Area_{\triangle PAC}$ has a maximum value,\\
The maximum value of $Area_{\triangle PAC}=\frac{4ac-{b}^{2}}{4a}=\frac{27}{8}$,\\
In the right triangle $\triangle AOC$, $AC=\sqrt{OC^{2}+OA^{2}}=\sqrt{3^{2}+3^{2}}=3\sqrt{2}$,\\
$\because Area_{\triangle PAC}=\frac{1}{2}AC\times PD$,\\
$\therefore \frac{27}{8}=\frac{1}{2}\times 3\sqrt{2}PD$,\\
$\therefore PD=\boxed{\frac{9\sqrt{2}}{8}}$,\\
Thus, when $Area_{\triangle PAC}$ reaches its maximum value of $\frac{27}{8}$, $PD$ has its maximum value of $\boxed{\frac{9\sqrt{2}}{8}}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ ($a\ne 0$) has its vertex at point Q$(2, -1)$ and intersects the $y$-axis at point C$(0, 3)$ and intersects the $x$-axis at points A and B (with point A to the right of point B). Point P is a moving point on the parabola, moving from point C towards point A (point P does not coincide with A). A perpendicular line $PD\perp$ to line $AC$ is drawn at point P. If point P coincides with point Q, point E is on the $x$-axis, and point F is on the parabola, is there a parallelogram with vertices A, P, E, F? If it exists, write down the coordinates of point F directly; if not, please explain the reason.

Solution: \textbf{Solution:} Solution: Exists, \boxed{F_1(2-\sqrt{2}, 1)}, \boxed{F_2(2+\sqrt{2}, 1)}.


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=BC$, point $C(2,0)$, and point $B(0,4)$. The graph of the inverse proportion function $y=\frac{k}{x} (k\ne 0)$ passes through point A. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} As shown in the figure, draw $AD\perp x$-axis at D, then $\angle ADC=90^\circ$, \\
$\because \angle ACB=90^\circ, AC=BC$, \\
$\therefore \angle BCO+\angle ACD=90^\circ$, \\
$\because \angle BCO+\angle CBO=90^\circ$, \\
$\therefore \angle ACD=\angle CBO$, \\
In $\triangle BOC$ and $\triangle CDA$, \\
$\left\{\begin{array}{l}
\angle BOC=\angle CDA=90^\circ \\
\angle ACD=\angle CBO \\
AC=BC
\end{array}\right.$, \\
$\therefore \triangle BOC\cong \triangle CDA\left(AAS\right)$, \\
$\because$ the coordinates of point C are $(2,0)$, and the coordinates of point B are $(0,4)$, \\
$\therefore OC=2, OB=4$, \\
$\therefore CD=OB=4, AD=OC=2$, \\
$\therefore OD=OC+CD=6$, \\
$\therefore$ the coordinates of point A are $(6,2)$, \\
Substituting the coordinates of point A into the inverse proportional function $y=\frac{k}{x}$, we get $k=2\times 6=12$, \\
$\therefore$ the analytic expression of the inverse proportional function is $\boxed{y=\frac{12}{x}}$;


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $AC=BC$, point $C(2, 0)$, point $B(0, 4)$, and the graph of the inverse proportion function $y=\frac{k}{x} (k\ne 0)$ passes through point A. The line $OA$ is translated upwards by $m$ units and then passes through the point $(3, n)$ on the graph of the inverse proportion function $y=\frac{k}{x} (k\ne 0)$. Find the value of $m$?

Solution:
\textbf{Solution:} Given the problem, \\
$\because$ $(3,n)$ lies on the curve of $y=\frac{12}{x}$, \\
$\therefore$ $n=4$, \\
Let the equation of line $OA$ be $y=k_1x$, \\
$\therefore$ $2=6k_1$, \\
$\therefore$ $k_1=\frac{1}{3}$, \\
$\therefore$ the equation of line $OA$ is $y=\frac{1}{3}x$, \\
$\therefore$ the equation of line $OA$ after being translated upwards by $m$ units is: $y=\frac{1}{3}x+m$, \\
$\because$ the line $y=\frac{1}{3}x+m$ goes through $(3,4)$, \\
$\therefore$ $4=\frac{1}{3}\cdot 3+m$, \\
Solving this, we find $m=\boxed{1}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-4$ passes through points $A(-8, 0)$ and $B(2, 0)$, and the line $x=-4$ intersects the $x$-axis at point $C$ and intersects the parabola at point $D$. What is the equation of this parabola?

Solution:
\textbf{Solution:} Since the parabola $y=ax^2+bx-4$ passes through points $A(-8, 0)$ and $B(2, 0)$,\\
let the equation of the parabola be $y=a(x+8)(x-2)=ax^2+6ax-16a$,\\
thus $-16a=-4$, solving this gives $a=\frac{1}{4}$,\\
therefore, the equation of the parabola is $\boxed{y=\frac{1}{4}x^2+\frac{3}{2}x-4}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-4$ passes through the points $A(-8, 0)$ and $B(2, 0)$, and the line $x=-4$ intersects the x-axis at point $C$ and intersects the parabola at point $D$. Point $P$ is on the parabola, and point $E$ is on the line $x=-4$. If the quadrilateral with vertices $A$, $O$, $E$, $P$ is a parallelogram, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Let point $P(m, \frac{1}{4}m^{2}+\frac{3}{2}m-4)$. When $AO$ is a side,\\
$\because$ the quadrilateral formed by vertices $A$, $O$, $E$, $P$ is a parallelogram, with point $A(0, -8)$,\\
$\therefore AO=PE=8$,\\
$\therefore |m+4|=8$,\\
Solving gives: $m_{1}=-12$, $m_{2}=4$,\\
When $m=4$, $\frac{1}{4}m^{2}+\frac{3}{2}m-4=\frac{1}{4}\times 16+\frac{3}{2}\times 4-4=6$;\\
When $m=-12$, $\frac{1}{4}m^{2}+\frac{3}{2}m-4=\frac{1}{4}\times 144+\frac{3}{2}\times (-12)-4=14$,\\
$\therefore$ the coordinates of point $P$ are $(-12, 14)$ or $(4, 6)$.\\
When $AO$ is a diagonal,\\
$\because$ point $A(-8, 0)$, point $C(-4, )$,\\
$\therefore$ points $A$, $O$ are symmetric about the line $x=-4$, the intersection points of line $x=-4$ with the parabola are $P$,\\
When $x=-4$, $\frac{1}{4}x^{2}+\frac{3}{2}x-4=\frac{1}{4}\times 16-\frac{3}{2}\times 4-4=-6$,\\
$\therefore$ the coordinates of point $P$ are $(-4, -6)$.\\
$\therefore$ the coordinates of point $P$ are $\boxed{(-12, 14)}$ or $\boxed{(4, 6)}$ or $\boxed{(-4, -6)}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-4$ passes through points $A(-8, 0)$ and $B(2, 0)$. The line $x=-4$ intersects the x-axis at point $C$ and intersects the parabola at point $D$. If the distances from points $B$, $D$, $C$ to the same line are $d_1$, $d_2$, $d_3$ respectively, is there a line $l$ such that $d_1=d_2=\frac{d_3}{2}$? If such a line exists, please directly write down the value of $d_3$; if it does not exist, please explain the reason.

Solution:
\textbf{Solution:} Exists. Connect BD, through point C draw $CH\perp BD$ at point H,

$\therefore$ Point C ($-4,0$), Point B ($2,0$), Point D ($-4,-6$),

$\therefore OC=4$, $OB=2$, $CD=6$,

$\therefore \triangle CDB$ is an isosceles right triangle,

$\therefore CH=CD\cdot\sin 45^\circ=6\times \frac{\sqrt{2}}{2}=3\sqrt{2}$,

$\because BD=2CH$,

$\therefore BD=6\sqrt{2}$,

$\because CO:OB=2:1$,

$\therefore$ The line $l_1$ passing through point O and parallel to BD satisfies the condition,

Draw line $BE\perp l_1$, $DF\perp l_1$ at point F, $CH$ intersects line $l_1$ at point G, outside $\triangle CDB$ draw line $l_2\parallel DB$, extending CH to intersect line $l_2$ at point G', making $CH=HG'$,

$\because BE=DF$, i.e., $d_1=d_2$,

$\therefore \frac{CG}{BE}=\frac{CO}{BO}=\frac{2}{1}$, $\frac{CG}{GH}=\frac{2}{1}$,

$\therefore \frac{d_3}{d_1}=\frac{2}{1}$,

$\therefore d_1=d_2=\frac{1}{2}d_3$,

$\therefore CG=\frac{2}{3}CH$,

$\therefore d_3=\frac{2}{3}\times 3\sqrt{2}=2\sqrt{2}$;

$\therefore d_3=CG' =2CH=6\sqrt{2}$;

As shown in the diagram, through points H, O draw line $l_3$, draw $BE\perp l_3$ at point E, $DF\perp l_3$ at point F, $CG\perp l_3$ at point G, draw $HI\perp x$-axis at point I,

$\therefore DH=BH$, $BE=DF$, i.e., $d_1=d_2$,

$\because CO:OB=2:1$,

$\therefore d_1:d_2=\frac{d_3}{2}$,

$\therefore HI=CI=\frac{1}{2}CB=3$,

$\therefore OI=4−3=1$,

$\therefore OH=\sqrt{HI^2+OI^2}=\sqrt{3^2+1^2}=\sqrt{10}$,

$\therefore S_{\triangle COH}=\frac{1}{2}\times 4\times 3=\frac{1}{2}\times \sqrt{10}d_3$,

Solving for this: $d_3=\frac{6\sqrt{10}}{5}$

$\therefore$ The value of $d_3$ is: $2\sqrt{2}$ or $6\sqrt{2}$ or $\boxed{\frac{6\sqrt{10}}{5}}$.


Question: As shown in the figure, a small ball M is thrown from point O on slope $OA$. The trajectory of the ball is a part of a parabola. A Cartesian coordinate system is established as shown in the figure. The slope can be described by the linear function $y=\frac{1}{2}x$. If the highest point reached by the ball has coordinates $(4, 8)$, find the expression for the parabola.

Solution:
\textbf{Solution:} Given that the highest point the ball reaches is at the coordinates $(4, 8)$,\\
we can assume the expression of the parabola to be $y=a(x-4)^2+8$. Since the parabola passes through the origin,\\
we have $0=a(0-4)^2+8$, solving for $a$ yields: $a=-\frac{1}{2}$,\\
thus, the expression of the parabola is $\boxed{y=-\frac{1}{2}(x-4)^2+8=-\frac{1}{2}x^2+4x}$.


Question: As shown in the figure, a small ball M is thrown from point O on the slope $OA$. The trajectory of the ball is part of a parabola. A Cartesian coordinate system is established as shown in the figure, and the slope can be described by the linear function $y=\frac{1}{2}x$. If the highest point that the ball reaches has the coordinates $(4, 8)$, answer the following question: There is a tree at point B on slope $OA$, where the x-coordinate of B is 2 and the height of the tree is 3.5. Can ball M fly over this tree? Explain your reasoning through calculation.

Solution:
\textbf{Solution:} By substituting $x_B=2$ into $y=\frac{1}{2}x$, we get: $y_B=\frac{1}{2}\times 2=1$,\\
$\therefore$ B(2, 1).\\
$\because$ The height of the tree is 3.5,\\
$\therefore$ the coordinates of the top of the tree are $(2, 4.5)$.\\
By substituting $x=2$ into $y=-\frac{1}{2}x^2+4x$, we get: $y=-\frac{1}{2}\times 2^2+4\times 2=6$,\\
$\therefore$ at this moment $M(2, 6)$,\\
$\therefore$ $y_M>y_{\text{top of the tree}}$,\\
$\therefore$ the ball M can fly over the tree, hence the answer is $\boxed{\text{Yes}}$.


Question: As shown in the figure, a small ball M is thrown from point O on slope $OA$. The trajectory of the ball is a part of a parabola. A Cartesian coordinate system is established as shown in the figure, and the slope can be characterized by the linear function $y=\frac{1}{2}x$. If the highest point reached by the ball is at coordinates $(4, 8)$, find the maximum height of ball M from the slope $OA$ during its flight.

Solution: \textbf{Solution:} Solve the system of equations $\begin{cases}y=-\frac{1}{2}x^2+4x\\ y=\frac{1}{2}x\end{cases}$, and we find: $\begin{cases}x_1=0\\ y_1=0\end{cases}$, $\begin{cases}x_2=7\\ y_2=\frac{7}{2}\end{cases}$. Therefore, $0\le x\le 7$. As shown in the figure, draw $MF\perp x$-axis at point F, intersecting $OA$ at point E. Let $E(t, \frac{1}{2}t)(0\le t\le 7)$, then $M(t, -\frac{1}{2}t^2+4t)$, hence $EM=y_M-y_E=-\frac{1}{2}t^2+4t-\frac{1}{2}t=-\frac{1}{2}(t-\frac{7}{2})^2+\frac{49}{8}$. Since $-\frac{1}{2}<0$ and $0\le t\le 7$, when $t=\frac{7}{2}$, $EM$ reaches its maximum value, which is \boxed{\frac{49}{8}} meters.


Question: As shown in the figure, in rectangle $OABC$, $OA=3$, $OC=4$. By respectively using the lines on which $OA$ and $OC$ lie as the $x$-axis and $y$-axis, we establish a Cartesian coordinate system. Let $D$ be a variable point on side $CB$ (not coinciding with $C$ or $B$). The graph of the inverse proportion function $y=\frac{k}{x}$ (where $k>0$) passes through point $D$ and intersects side $BA$ at point $E$. Line $DE$ is drawn. When point $D$ moves to the midpoint of $BC$, find the value of $k$?

Solution:
\textbf{Solution:} Since $OA=3$, $OC=4$, and quadrilateral $OABC$ is a rectangle,\\
it follows that $BC=OA=3$, and the coordinates of point $B$ are $(3,4)$. Since point $D$ is the midpoint of side $BC$,\\
it follows that $CD=\frac{1}{2}BC=\frac{3}{2}$\\
Therefore, the coordinates of point $D$ are  $\left(\frac{3}{2}, 4\right)$\\
Furthermore, since point $D$ lies on the graph of the inverse proportion function $y=\frac{k}{x}(k>0)$,\\
it follows that $k=\frac{3}{2}\times 4=\boxed{6}$


Question: As shown in the figure, in rectangle $OABC$, we have $OA=3$ and $OC=4$. The lines containing $OA$ and $OC$ are taken as the $x$-axis and $y$-axis, respectively, establishing a Cartesian coordinate system. Let $D$ be a variable point on side $CB$ (not coinciding with $C$ or $B$), and the graph of the inverse proportion function $y=\frac{k}{x}$ $(k>0)$ passes through point $D$ and intersects side $BA$ at point $E$. Construct line $DE$. Find the value of $\frac{BD}{BE}$?

Solution:
Solution: Since points D and E lie on the graph of the inverse proportion function $y=\frac{k}{x}$ ($k>0$),\\
it follows that the coordinates of D are $\left(\frac{k}{4}, 4\right)$, and the coordinates of E are $\left(3, \frac{k}{3}\right)$.\\
Given also that the coordinates of point B are (3, 4),\\
it follows that $BD=3-\frac{k}{4}$ and $BE=4-\frac{k}{3}$,\\
therefore, $\frac{BD}{BE}=\frac{3-\frac{k}{4}}{4-\frac{k}{3}}=\boxed{\frac{3}{4}}$.


Question: As shown in the figure, in rectangle $OABC$, $OA=3$, $OC=4$. The lines containing $OA$ and $OC$ are taken as the $x$-axis and the $y$-axis, respectively, to establish a Cartesian coordinate system. Point $D$ is a moving point on side $CB$ (not coinciding with $C$ or $B$), and the graph of the inverse proportion function $y=\frac{k}{x}$ ($k>0$) passes through point $D$ and intersects side $BA$ at point $E$. Line $DE$ is drawn. Connect $DA$. When the area of $\triangle DAE$ is $\frac{4}{3}$, what is the value of $k$?

Solution:
\textbf{Solution:} From equation (2), we know that $AE=\frac{k}{3}$ and $BD=3-\frac{k}{4}$,\\
therefore, the area of $\triangle DAE$ is ${S}_{\triangle DAE}=\frac{1}{2}AE\cdot BD=\frac{1}{2}\times \frac{k}{3}\times \left(3-\frac{k}{4}\right)=\frac{4}{3}$,\\
Simplifying, we get: ${k}^{2}-12k+32=0$,\\
Solving, we find ${k}_{1}=\boxed{4}$ and ${k}_{2}=\boxed{8}$, therefore, when the area of $\triangle DAE$ is $\frac{4}{3}$, the value of $k$ is \boxed{4} or \boxed{8}.


Question: Given that the parabola $y=ax^2-3ax-4a$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$), and intersects the negative half of the y-axis at point $C$, and $P$ is a point on the parabola in the fourth quadrant. If the area of $\triangle ABC$ is $5$, what is the value of $a$?

Solution:
\textbf{Solution:} Set $y=0$, we obtain $ax^{2}-3ax-4a=0$, solving this gives: $x_{1}=-1$, $x_{2}=4$.\\
Since $y=ax^{2}-3ax-4a$ intersects with the x-axis at points $AB$ (A on the left and B on the right), and intersects the y-axis at point C,\\
Therefore, $A(-1,0)$, $B(4,0)$, $C(0,-4a)$, and $AB=5$,\\
Since the area ${S}_{\triangle ABC}=5$,\\
${S}_{\triangle ABC}=\frac{1}{2}\times AB\times OC=\frac{1}{2}\times 5\times 4a=5$,\\
Solving gives: $a=\boxed{\frac{1}{2}}$;


Question: Given the parabola $y=ax^{2}-3ax-4a$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the negative y-axis at point $C$. Let $P$ be a point on the parabola in the fourth quadrant. If $a=1$, a line through point $P$ is drawn perpendicular to the x-axis, intersecting line segment $BC$ at point $Q$. Find the maximum value of the line segment $PQ$ and the coordinates of point $P$ at this time.

Solution: \textbf{Solution:} When $a=1$, the parabola is $y=x^{2}-3x-4$. By substituting the coordinates of points $B(4,0)$ and $C(0,-4)$ into the expression of the linear function, we obtain the equation of line $BC$ as: $y=x-4$. Let point $P$ be $(t,t^{2}-3t-4)$, then point $Q$ is $(t,t-4)$, $\therefore PQ=(t-4)-(t^{2}-3t-4)=-t^{2}+4t=-(t-2)^{2}+4$, $\therefore$ when $t=2$, $PQ$ reaches its maximum value $\boxed{4}$, at which point $P$ is $\boxed{(2,-6)}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2-3ax-4a$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the negative half of the y-axis at point $C$. Let $P$ be a point on the parabola in the fourth quadrant. The line $AP$ intersects the y-axis at point $M$, and the line $BP$ intersects the y-axis at point $N$. Find the value of $\frac{4OM+ON}{OC}$.

Solution: \textbf{Solution:} Given from (1) that points $A(-1,0)$, $B(4,0)$, and $C(0,-4a)$. Let point $P(m,am^{2}-3am-4a)$, by substituting the coordinates of points P and A into the expression of a linear function and solving, we find the equation of line $PA$ to be: $y=a(m-4)x+a(m-4)$. Therefore, $OM=-a(m-4)$. Similarly, the equation of line $BP$ is $y=a(m+1)x-4a(m+1)$, $ON=4a(m+1)$. Hence, $4OM+ON=-4a(m-4)+4a(m+1)=20a$. Since $C(0,-4a)$, it follows that $OC=4a$. Thus, $\frac{4OM+ON}{OC}=\frac{20a}{4a}=\boxed{5}$.


Question: As shown in the figure, it is known that the parabola $y=x^{2}+5x+4$ intersects the $x$-axis at point $A$ (point $A$ is to the left of the axis of symmetry of this parabola), and intersects the $y$-axis at point $B$. The line segment $AB$ intersects the axis of symmetry of the parabola at point $P$. $OA=OB$. Find the coordinates of point $P$.

Solution:
\[ \textbf{Solution:} \text{ Let the equation of line AB be } y=kx+b, \]
\[ \text{then } \begin{cases} b=4 \\ 4k+b=0 \end{cases} \]
\[ \text{Solving yields } \begin{cases} k=-1 \\ b=4 \end{cases} \]
\[ \therefore \text{ the equation of line AB is } y=-x+4, \]
\[ \because \text{ the axis of symmetry of the parabola is the line } x=-\frac{5}{2}, \]
\[ \therefore \text{ when } x=-\frac{5}{2}, y=-\frac{5}{2}+4=\frac{3}{2}, \]
\[ \therefore \text{ the coordinates of point } P \text{ are } \boxed{\left(-\frac{5}{2}, \frac{3}{2}\right)}. \]


Question: As shown in the figure, it is known that in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b$ ($k\neq 0$) passes through points A and $B(-1,0)$, and the graph of the inverse proportion function $y=\frac{6}{x}$ also passes through point A, whose abscissa is 2. Find the analytical expression of the linear function.

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y=\frac{6}{x}$ passes through point A, and the abscissa of point A is 2,\\
hence $y_{A}=\frac{6}{2}=3$, that is, $A(2,3)$.\\
Since the graph of the linear function $y=kx+b(k\ne 0)$ passes through points A, $B(-1,0)$,\\
thus $\left\{\begin{array}{l}3=2k+b \\ 0=-k+b \end{array}\right.$,\\
solving this, we get: $\left\{\begin{array}{l}k=1 \\ b=1 \end{array}\right.$,\\
therefore the explicit formula of the linear function is $\boxed{y=x+1}$;


Question: As shown in the figure, it is known that in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b(k\ne 0)$ passes through points A and $B(-1,0)$, the graph of the inverse proportion function $y=\frac{6}{x}$ also passes through point A, and the abscissa of point A is 2. Point C is a point on the positive half of the x-axis, connect $AC$, $\tan\angle ACB=\frac{3}{4}$. A line through point C is drawn perpendicular to the x-axis and intersects the graphs of the inverse proportion function $y=\frac{6}{x}$ and the linear function $y=kx+b(k\ne 0)$ at points D and E, respectively. Find the coordinates of points D and E.

Solution:
\textbf{Solution:} As shown in the figure, draw $AH \perp x$-axis at point H.

$\therefore \tan\angle ACB = \frac{AH}{CH} = \frac{3}{4}$.

$\because A(2,3)$,

$\therefore AH=3$, $OH=2$

$\therefore CH=4$,

$\therefore OC=OH+CH=6$,

$\therefore C(6,0)$.

$\because x_C=x_D=x_E=6$.

$\therefore y_D=\frac{6}{x_D}=1$, $y_E=x_E+1=7$,

$\therefore D(6,1)$, $E(6,7)$;

Hence, the coordinates of points D and E are $D=\boxed{(6,1)}$, $E=\boxed{(6,7)}$.


Question: As shown in the diagram, it is known that in the Cartesian coordinate system $xOy$, the graph of the linear function $y=kx+b(k\ne 0)$ passes through points A and $B(-1,0)$, and the graph of the hyperbolic function $y=\frac{6}{x}$ also passes through point A, where the abscissa of point A is 2. Under this condition, when connecting $AD$, is there a point F on the graph of the linear function $y=kx+b(k\ne 0)$ such that triangles $\triangle EAD$ and $\triangle ECF$ are similar? If such a point exists, please write down the coordinates of point F directly; if not, please explain why.

Solution: \textbf{Solution:} Exist, the coordinates of point F are $\boxed{\left(\frac{4}{3},\frac{7}{3}\right)}$.


Question: As shown in the figure, the parabola $y=ax^2-a$ (where $a>0$) intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the y-axis at point $C$. Given that $OC = 2OA$, what are the coordinates of points $A$ and $B$?

Solution:
\textbf{Solution:} According to the given information, we have $y=0$, which implies $ax^2-a=0$, \\
thus $a(x+1)(x-1)=0$, \\
solving this we get: $x=-1$ or $x=1$, \\
thus, we have the points \boxed{A(-1, 0)}, \boxed{B(1, 0)}.


Question: As shown in the figure, the parabola $y=ax^{2}-a$ ($a>0$) intersects the x-axis at points $A$ and $B$ ($A$ is to the left of $B$), and intersects the y-axis at point $C$, with $OC=2OA$. What is the analytical expression of the parabola?

Solution:
\textbf{Solution:}
Given $A(-1, 0)$, \\
$\therefore OA=1$, \\
$\therefore OC=2OA=2$, \\
$\therefore C(0, -2)$, \\
Substituting into the parabola $y=ax^{2}-a(a>0)$, we get: $-a=-2$, \\
$\therefore a=2$, \\
$\therefore$ The equation of the parabola is $y=2x^{2}-2=\boxed{2x^{2}-2}$.


Question: As shown in the figure, the parabola $y=ax^2-a$ ($a>0$) intersects the x-axis at points $A$ and $B$ (point $A$ is to the left of point $B$), and intersects the y-axis at point $C$, with $OC=2OA$. Is there a point $P$ on the parabola such that the incenter of $\triangle PAC$ lies on the x-axis? If it exists, find the coordinates of point $P$; if not, please explain why.

Solution:
\textbf{Solution:} Existence is proven,\\
Assume there exists a point $P$ such that the intersection of the angle bisectors of the three interior angles of $\triangle PAC$ lies on the $x$-axis, then the $x$-axis is the angle bisector of $\angle PAC$.\\
$\therefore$ The reflection point of $C$ about the $x$-axis must lie on line $PA$. Denote it as $C'$,\\
$\because$ $C(0, -2)$,\\
$\therefore$ $C'(0, 2)$,\\
$\therefore$ Line $AP$ passes through $A(-1, 0), C'(0, 2)$,\\
Let the equation of line $AP$ be $y=kx+2$,\\
$\therefore$ $-k+2=0$, solving yields $k=2$,\\
$\therefore$ The equation of line $AP$ is $y=2x+2$,\\
$\because$ Line $AP$ intersects the quadratic function $y=2x^2-2$ at point $P$,\\
$\therefore$ $2x+2=2x^2-2$,\\
Solving gives: $x=2$ or $x=-1$,\\
When $x=2$, $y=6$,\\
When $x=-1$, $y=0$, which corresponds to point $A$,\\
$\therefore$ There exists a point $P$ such that the intersection of the angle bisectors of the three interior angles of $\triangle PAC$ lies on the $x$-axis, and the coordinates of point $P$ are $\boxed{(2,6)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $C_1\colon y=-\frac{1}{4}x^2+bx+c$ passes through the points $(-3,0)$ and $(3,3)$, and intersects the y-axis at point C. What is the equation of the parabola $C_1$?

Solution:
\textbf{Solution:} Since the graph of $y=-\frac{1}{4}x^2+bx+c$ passes through the points $(-3,0)$ and $(3,3)$,\\
we have
\[
\begin{cases}
-\frac{1}{4}(-3)^2+3b+c=0\\
-\frac{1}{4}3^2+3b+c=3
\end{cases},
\]
from which we deduce
\[
\begin{cases}
b=\frac{1}{2}\\
c=\frac{15}{4}
\end{cases},
\]
Therefore, the equation of the parabola $C_1$ is $\boxed{y=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{15}{4}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $C_1\colon y=-\frac{1}{4}x^2+bx+c$ passes through the points $(-3,0)$ and $(3,3)$, and intersects the y-axis at point C. The parabola $C_1$ is first translated 3 units to the left and then 2 units downward to obtain the parabola $C_2$. The vertex of parabola $C_2$ is D. What is the area of $\triangle ODC$?

Solution:
Solution: Since $y=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{15}{4}=-\frac{1}{4}(x-1)^2+4$,\\
the vertex of parabola $C_1$ is located at $(1,4)$, and it intersects the y-axis at point $C(0,\frac{15}{4})$,\\
since moving parabola $C_1$ 3 units to the left and 2 units down yields parabola $C_2$,\\
the vertex $D$ of parabola $C_2$ has coordinates $(-2,2)$,\\
therefore, the area of $\triangle ODC$ is $\frac{1}{2}\times 2\times \frac{15}{4}=\boxed{\frac{15}{4}}$.


Question: As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points A and B, with point A to the left of point B, and intersects the y-axis at point C. Point D is a moving point on the parabola. Find the coordinates of points A, B, and C.

Solution:
\textbf{Solution:} Substitute $y=0$ into $y=2x^{2}-4x-6$, \\
we get $2x^{2}-4x-6=0$. \\
Solving this, we find $x_{1}=-1$, $x_{2}=3$. \\
Therefore, the coordinates of point A are $\boxed{\left(-1, 0\right)}$, and the coordinates of point B are $\boxed{\left(3, 0\right)}$. \\
Substitute $x=0$ into $y=2x^{2}-4x-6$, we get $y=-6$. \\
Therefore, the coordinates of point C are $\boxed{\left(0, -6\right)}$.


Question: As shown in the figure, the parabola $y=2x^2-4x-6$ intersects the x-axis at points $A$ and $B$, with point $A$ to the left of point $B$, and intersects the y-axis at point $C$. Point $D$ is a moving point on the parabola. As shown in figure 2, when point $D$ is in the fourth quadrant, connect $BD$, $CD$, and $BC$ to form $\triangle BCD$. When the area of $\triangle BCD$ is maximized, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} Let the coordinates of point D be $\left(m, 2m^{2}-4m-6\right)$.

As shown in the figure, draw $DH\perp x$ axis at point H and $DG\perp y$ axis at point G, then connect $OD$.

Therefore, $DG=m$, $DH=-2m^{2}+4m+6$.

Because the coordinates of point B are $\left(3, 0\right)$ and the coordinates of point C are $\left(0, -6\right)$.

Therefore, $OB=3$, $OC=6$.

Because $S_{\triangle BCD}=S_{\triangle OCD}+S_{\triangle OBD}-S_{\triangle OBC}$,

Therefore, $S_{\triangle BCD}=\frac{OC\cdot DG}{2}+\frac{OB\cdot DH}{2}-\frac{OC\cdot OB}{2}$

$S_{\triangle BCD}=\frac{6m}{2}+\frac{3(-2m^{2}+4m+6)}{2}-\frac{6\times 3}{2}$

Upon simplification, we get $S_{\triangle BCD}=-3\left(m-\frac{3}{2}\right)^{2}+\frac{27}{4}$.

Because $-3<0$,

Therefore, when $m=\frac{3}{2}$, the maximum area of $\triangle BCD$ is $\frac{27}{4}$.

Therefore, $2m^{2}-4m-6=2\times \left(\frac{3}{2}\right)^{2}-4\times \frac{3}{2}-6=2\times \frac{9}{4}-6-6=-\frac{15}{2}$.

Therefore, the coordinates of point D are $\boxed{\left(\frac{3}{2}, -\frac{15}{2}\right)}$.


Question: As shown in the figure, it is known that the graph of the quadratic function $y=-x^2+bx+c$ passes through two points $A(-2, -1)$ and $B(0, 7)$. What is the analytic expression of the parabola and its axis of symmetry?

Solution:
\textbf{Solution:} Let the graph of the quadratic function $y=-x^2+bx+c$ pass through points $A(-2, -1)$, and $B(0, 7)$.

Hence, $\left\{\begin{array}{l}
-1=-4-2b+c \\
c=7
\end{array}\right.$,

Solving this system, we get: $\left\{\begin{array}{l}
b=2 \\
c=7
\end{array}\right.$,

Therefore, $y=-x^2+2x+7$,

$=-\left(x^2-2x\right)+7$,

$=-\left[\left(x^2-2x+1\right)-1\right]+7$,

$=-(x-1)^2+8$,

Hence, the axis of symmetry is the line $x=\boxed{1}$.


Question: As shown in the figure, it is known that the graph of the quadratic function $y=-x^2+bx+c$ passes through points $A(-2, -1)$ and $B(0, 7)$. For what values of $x$ is $y>0$?

Solution:
\textbf{Solution}: When $y=0$,\\
$0=-(x-1)^2+8$,\\
$\therefore x-1=\pm 2\sqrt{2}$,\\
${x}_{1}=1+2\sqrt{2}$, ${x}_{2}=1-2\sqrt{2}$,\\
$\therefore$ the intersection points of the parabola and the $x$-axis are: $(1-2\sqrt{2}, 0)$, $(1+2\sqrt{2}, 0)$,\\
$\therefore$ when \boxed{1-2\sqrt{2}<x<1+2\sqrt{2}}, $y>0$.


Question: As illustrated, it is known that the graph of the quadratic function $y=-x^2+bx+c$ passes through points $A(-2, -1)$ and $B(0, 7)$. On the upper side of the x-axis, a line $l$ is drawn parallel to the x-axis, intersecting the parabola at points $C$ and $D$ (with point $C$ being on the left side of the axis of symmetry). Perpendiculars are drawn from points $C$ and $D$ to the x-axis, with foot points at $F$ and $E$, respectively. When the rectangle $CDEF$ becomes a square, what are the coordinates of point $C$?

Solution:
\textbf{Solution:} When the rectangle $CDEF$ is a square,\\
let the coordinates of point $C$ be $(x, -x^2+2x+7)$,\\
$\therefore$ the coordinates of point $D$ are $(-x^2+3x+7, -x^2+2x+7)$,\\
$\because$ the axis of symmetry is the line $x=1$, and the distance from $D$ to the axis of symmetry is equal to the distance from $C$ to the axis of symmetry,\\
$\therefore -x^2+3x+7-1=-x+1$,\\
solving this gives $x_1=-1$, $x_2=5$ (discard as it does not meet the condition),\\
when $x=-1$, $-x^2+2x+7=4$,\\
$\therefore$ the coordinates of point $C$ are: $\boxed{(-1, 4)}$.


Question: As shown in the figure, a parabola $y=ax^2+bx+c$ ($a\neq 0$) with the axis of symmetry $x=-1$ intersects the x-axis at points $A$ and $B$ $(-3, 0)$, and the point $(2, 5)$ lies on the parabola $y=ax^2+bx+c$. Determine the equation of the parabola.

Solution:
\textbf{Solution}: By the symmetry axis of the parabola being $x=-1$ and point $A(-3, 6)$ being on the parabola, we get the following system of equations:
\[
\begin{cases}
9a-3b+c=6\\
4a+2b+c=5\\
-\frac{b}{2a}=-1
\end{cases}
\]
Solving this, we obtain:
\[
\begin{cases}
a=3\\
b=2\\
c=-3
\end{cases}
\]
$\therefore$ The equation of the parabola is: \boxed{y=3x^2+2x-3}.


Question: Given that the graph of the parabola $y=ax^2+bx+3$ intersects the x-axis at point A and point $B(1,0)$, and intersects the y-axis at point C. Connect $AC$. Let a moving point D be on segment $AC$. Draw a perpendicular to the x-axis from point D, which intersects the parabola at point E and intersects the x-axis at point F. Given $AB=4$, let the x-coordinate of point D be m. Find the equation of the parabola?

Solution:
Solution: Since point $B(1,0)$ and $AB=4$, \\
therefore $A(-3,0)$, \\
substituting $B(1,0)$ and $A(-3,0)$ into $y=ax^2+bx+3$, \\
therefore $\begin{cases}
a+b+3=0\\
9a-3b+3=0
\end{cases}$, \\
thus $\begin{cases}
a=-1\\
b=-2
\end{cases}$, \\
therefore the equation of the parabola is $\boxed{y=-x^2-2x+3}$;


Question: As shown in the figure, it is known that the parabola $y = ax^2 + bx + 3$ intersects the x-axis at points A and $B(1,0)$, and intersects the y-axis at point C. Line segment $AC$ is drawn, and a moving point D moves on segment $AC$. A perpendicular line to the x-axis is drawn from point D, intersects the parabola at point E, and intersects the x-axis at point F. It is given that $AB = 4$. Let the x-coordinate of point D be m. Lines $AE$ and $CE$ are drawn. When the area of $\triangle ACE$ is maximized, find the maximum area of $\triangle ACE$ and the coordinates of point D.

Solution:
\textbf{Solution:} Suppose the equation of line AC is $y=k'x+b'$,

$\therefore \left\{\begin{array}{l}
-3k'+b'=0 \\
b'=3
\end{array}\right.$, solving this we get: $\left\{\begin{array}{l}
k'=1 \\
b'=3
\end{array}\right.$,

$\therefore$ the equation of line AC is $y=x+3$,

$\therefore D(m, m+3), E(m, -m^2-2m+3)$,

$\therefore DE=-m^2-3m$,

$\therefore {S}_{\triangle ACE}=\frac{1}{2}\times 3\times (-m^2-3m)=-\frac{3}{2}(m+\frac{3}{2})^2+\frac{27}{8}$,

$\therefore$ when $m=-\frac{3}{2}$, the maximum value of ${S}_{\triangle ACE}$ is $\boxed{\frac{27}{8}}$,

$\therefore \boxed{D\left(-\frac{3}{2}, \frac{3}{2}\right)}$;


Question: Given the parabola $y=ax^2+bx+3$ intersects the x-axis at points A and $B(1,0)$, and the y-axis at point C. Connect $AC$. A moving point D is on the line segment $AC$. Draw a perpendicular line from D to the x-axis, which intersects the parabola at point E and the x-axis at point F. Given $AB=4$, let the x-coordinate of point D be $m$. When $m=-2$, is there a point Q in the plane such that the quadrilateral formed by vertices B, C, E, Q is a parallelogram? If such a point exists, please provide the coordinates of point Q directly; if not, please explain why.

Solution: \textbf{Solution:} When point $Q$ is at $\boxed{(3, 0)}$, $\boxed{(-1, 0)}$, or $\boxed{(-3, 6)}$, the quadrilateral formed by vertices B, C, E, and Q is a parallelogram.


Question: The parabola $y=ax^2+bx+4$ intersects the x-axis at points $A(-4, 0)$ and $B(1, 0)$, and intersects the y-axis at point C. Point P is a moving point on the parabola above the line $AC$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Since the parabola $y=ax^2+bx+4$ intersects the x-axis at points $A(-4, 0)$ and $B(1, 0)$, \\
we have $\left\{\begin{array}{l}
16a+4b+4=0 \\
a+b+4=0
\end{array}\right.$. \\
Solving this system, we find $\left\{\begin{array}{l}
a=-1 \\
b=-3
\end{array}\right.$. \\
Therefore, the quadratic function is $\boxed{y=-x^2-3x+4}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+4$ intersects the x-axis at points $A(-4, 0)$ and $B(1, 0)$, and intersects the y-axis at point C. Point P is a moving point on the parabola above line $AC$. Draw $PE\perp AC$ at point E. Find the maximum value of $\frac{\sqrt{2}}{2}PE$ and the coordinates of point P at this time.

Solution:
\textbf{Solution:} Given that the quadratic function $y=-x^2-3x+4$ intersects with the y-axis at point C,\\
hence the coordinates of point C are $(0, 4)$,\\
let the equation of line $AC$ be $y=kx+4$,\\
since line $AC$ passes through point $A(-4, 0)$,\\
therefore $0=-4k+4$,\\
solving for $k$ gives $k=1$,\\
thus the equation of line $AC$ is $y=-x+4$,\\
draw $PF\perp x$-axis at point F, intersecting line $AC$ at point D,\\
let point $P(x, -x^2-3x+4)$, then $D(x, x+4)$,\\
hence $PD=-x^2-4x=-(x+2)^2+4$,\\
thus when $x=-2$, $PD$ is maximized, with the maximum value being $4$.\\
Since $A(-4, 0)$, $C(0, 4)$,\\
therefore $OA=OC$,\\
hence $\angle OAC=45^\circ$,\\
since $PF\perp x$-axis,\\
thus $\angle ADF=\angle PDE=45^\circ$,\\
since $PE\perp AC$,\\
therefore $\triangle PDE$ is an isosceles right triangle,\\
thus $\sqrt{2}PE=PD$,\\
hence $\frac{\sqrt{2}}{2}PE=\frac{1}{2}PD$,\\
therefore the maximum value of $\frac{\sqrt{2}}{2}PE$ is $\frac{1}{2}\times 4=2$,\\
at this moment, the coordinates of point P are $(-2, 6)$,\\
therefore, the maximum value of $\frac{\sqrt{2}}{2}PE$ is $\boxed{2}$, with the coordinates of point P being $\boxed{(-2, 6)}$.


Question: As shown in the diagram, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Connect $BC$. Find the analytical expression of the parabola.

Solution:
\textbf{Solution:} Substitute points $A(-1, 0)$ and $B(3, 0)$ into $y=ax^{2}+bx+3$,\\
we get $\begin{cases}a-b+3=0\\ 9a+3b+3=0\end{cases}$,\\
solving this yields $\begin{cases}a=-1\\ b=2\end{cases}$,\\
$\therefore \boxed{y=-x^{2}+2x+3}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Connect $BC$. Point P is a point on the parabola above line $BC$, connect $PB$ and $PC$. Find the maximum area of $\triangle PBC$ and the coordinates of point P at this time.

Solution:
\textbf{Solution:} As shown in the figure, draw $PF\perp AB$ at $F$, intersecting $BC$ at $E$,\\
since the intersection of the parabola $y=-x^2+2x+3$ with the y-axis is point $C$,\\
thus $C(0, 3)$,\\
suppose the equation of line $BC$ is $y=kx+b\ (k\ne 0)$,\\
substituting $B(3, 0)$, $C(0, 3)$ into the equation yields $\begin{cases}3k+b=0\\ b=3\end{cases}$,\\
solving this, we get $\begin{cases}k=-1\\ b=3\end{cases}$,\\
therefore, the equation of line $BC$ is: $y=-x+3$,\\
let point $P(t, -t^2+2t+3)$,\\
thus $E(t, -t+3)$,\\
thus $PE=-t^2+3t$,\\
thus the area of $\triangle PBC=\frac{1}{2}\times (-t^2+3t)\times 3=-\frac{3}{2}\left(t-\frac{3}{2}\right)^2+\frac{27}{8}$,\\
thus, when $t=\frac{3}{2}$, the maximum area of $\triangle PBC$ is $\boxed{\frac{27}{8}}$,\\
thus, point $\boxed{P\left(\frac{3}{2}, \frac{15}{4}\right)}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Line $BC$ is drawn. When the parabola $y=ax^2+bx+3$ is translated 1 unit to the right, a new parabola is obtained. Point M is on the axis of symmetry of the new parabola and point N is a moving point on the new parabola. If the quadrilateral formed by points M, N, B, and C is a parallelogram, directly write down the coordinates of point M.

Solution: \textbf{Solution:} The coordinates of point $M$ are $\boxed{(2, -8)}$, $\boxed{(2, -2)}$, or $\boxed{(2, 0)}$.


Question: As shown in the figure, it is known that the graph of the parabola $y=x^2+bx+c$ intersects the x-axis at point $A(-1,0)$ and intersects the y-axis at point $C(0,3)$. Find the analytical expression of the parabola and the coordinates of the other intersection point $B$ with the x-axis.

Solution:
Solution: Substituting points $A(-1,0)$ and $C(0,-3)$ into the parabola $y=x^{2}+bx+c$ yields the system of equations
\[
\left\{
\begin{array}{l}
1-b+c=0\\
c=-3
\end{array}
\right.
\]
Solving the system of equations, we get:
\[
\left\{
\begin{array}{l}
b=-2\\
c=-3
\end{array}
\right.
\]
Hence, the function equation is $y=x^{2}-2x-3$. When $y=0$, $x^{2}-2x-3=0$, solving this gives $x_{1}=-1$, $x_{2}=3$;\\
The coordinate of another intersection point $B$ is $\boxed{(3,0)}$.


Question: As shown in the figure, the graph of the parabola $y=x^2+bx+c$ intersects the x-axis at one point $A(-1,0)$ and the y-axis at point $C(0,3)$. Based on the graph, for what values of $x$ is $y<0$?

Solution: \textbf{Solution:} It can be concluded from the graph that when $-1<x<3$, $y<0$. Therefore, when $x$ takes $\boxed{-1<x<3}$, $y<0$.


Question: Given the parabola $y=x^2+bx+c$ as shown in the figure, it intersects the x-axis at point $A(-1,0)$ and the y-axis at point $C(0,3)$. On the axis of symmetry of the parabola, there is a moving point $P$. Find the coordinates of point $P$ when the value of $PA+PC$ is minimized.

Solution:
\textbf{Solution:} As shown in the diagram, construct the axis of symmetry and draw line BC. The intersection point P is the point we are seeking. Suppose line BC is $y=kx+b$. Substituting point $(3,0)$ and $C(0,-3)$ into it, we obtain
\[
\left\{
\begin{array}{l}
3k+b=0\\
b=-3
\end{array}
\right.
\]
Solving this, we get
\[
\left\{
\begin{array}{l}
k=1\\
b=-3
\end{array}
\right.
\]
$\therefore$ The equation of line BC is: $y=x-3$, and because $y=x^2-2x-3=(x-1)^2-4$, thus the axis of symmetry of the parabola is the line $x=1$. Substituting $x=1$ into $y=x-3$ yields $y=-2$. $\therefore$ The coordinates of point $P$ are $\boxed{(1,-2)}$.


Question: As shown in the figure, it is known that the parabola passes through three points $A(4,0)$, $B(1,0)$, and $C(0,-2)$. Find the equation of the parabola.

Solution: \textbf{Solution:} Let the equation of the parabola be $y=a(x-4)(x-1)$. Substituting the coordinates of point $C(0, -2)$ into the equation gives: $-2=a(0-4)(0-1)$, solving for $a$ yields $a=-\frac{1}{2}$, $\therefore y=-\frac{1}{2}(x-4)(x-1)=-\frac{1}{2}x^{2}+\frac{5}{2}x-2$. Hence, the equation of the parabola is $\boxed{y=-\frac{1}{2}x^{2}+\frac{5}{2}x-2}$.


Question: As shown in the figure, it is known that the parabola passes through points A$(4, 0)$, B$(1, 0)$, and C$(0, -2)$. Is there a point D on the parabola above line AC for which the area of $\triangle DCA$ is maximized? If such a point exists, find the coordinates of point D and the maximum area of $\triangle DCA$; if it does not exist, please explain why.

Solution:
\textbf{Solution:} As shown in the figure, connect AC and DC. Assume there exists a point D on the parabola, and connect AD and CD. Draw $DE \perp x$ axis from point D and let it intersect line AC at point E. Assume the expression for line AC is $y=kx+b$ ($k\neq 0$), substituting A(4, 0) and C(0, -2) into this expression, we get:
\[
\begin{cases}
0=4k+b\\
-2=b
\end{cases}
\]
From which, we solve:
\[
\begin{cases}
k=\frac{1}{2}\\
b=-2
\end{cases}
\]
$\therefore$ The equation of line AC is $y=\frac{1}{2}x-2$. Assume point D has coordinates $\left(x, -\frac{1}{2}x^{2}+ \frac{5}{2}x-2\right)$, then coordinates of point E are $\left(x, \frac{1}{2}x-2\right)$,\\
$\therefore$ The area of $\triangle DCA$ equals to $S_{\triangle DCA}=S_{\triangle DCE}+S_{\triangle DAE}=\frac{1}{2}\cdot DE\cdot x_{E}+\frac{1}{2}\cdot DE\cdot(x_{A}-x_{E})=\frac{1}{2}\cdot DE\cdot x_{A}=\frac{1}{2}\cdot DE\cdot 4=2DE$,\\
$\because$ $DE=\left(-\frac{1}{2}x^{2}+\frac{5}{2}x-2\right)-\left(\frac{1}{2}x-2\right)=-\frac{1}{2}x^{2}+2x$,\\
$\therefore$ The area of $\triangle DCA$ equals $2DE=2\left(-\frac{1}{2}x^{2}+2x\right)=-x^{2}+4x=-\left(x-2\right)^{2}+4$,\\
$\therefore$ When $x=2$, the area of $\triangle DCA$ reaches its maximum value, which is $\boxed{4}$,\\
Substituting $x=2$ into the equation $y=-\frac{1}{2}x^{2}+\frac{5}{2}x-2$, we solve and get $y=1$,\\
$\therefore$ Point D is at \boxed{D(2,1)}.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ passes through point $A(-4,0)$ and point $C(0,-4)$, with another intersection point with the x-axis being B. What is the analytical expression of this parabola?

Solution:
\textbf{Solution}: Since the parabola $y={x}^{2}+bx+c$ passes through points $A\left(-4,0\right)$ and $C\left(0,-4\right)$,\\
thus $\begin{cases}16-4b+c=0\\ c=-4\end{cases}$,\\
therefore $\begin{cases}b=3\\ c=-4\end{cases}$,\\
hence the equation of the parabola is $\boxed{y={x}^{2}+3x-4}$;


Question: As shown in the figure, the parabola $y=x^{2}+bx+c$ passes through points $A(-4,0)$ and $C(0,-4)$, with another intersection point with the x-axis being $B$. If the vertex is $D$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} Since the equation of the parabola is $y=x^{2}+3x-4=\left(x+\frac{3}{2}\right)^{2}-\frac{25}{4}$,\\
the coordinates of the vertex D of the parabola are $\boxed{\left(-\frac{3}{2},-\frac{25}{4}\right)}$.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ passes through point $A(-4,0)$ and point $C(0,-4)$, with another intersection point with the x-axis being $B$. When $y>0$, find the range of values for $x$.

Solution:
\textbf{Solution:} Let $y=0$, then $x^{2}+3x-4=0$, \\
solving this yields $x_{1}=-4$ and $x_{2}=1$, \\
$\therefore$ the coordinates of point B are $\left(1,0\right)$, \\
$\therefore$ from the graph of the function, it is known that when $y>0$, \boxed{x<-4} or \boxed{x>1}.


Question: As shown in the figure, the line $y_1=-x+3$ intersects the x-axis and y-axis at points $B$ and $C$ respectively. The parabola $y_2=ax^2+bx+c$ passing through points $B$ and $C$ has another intersection point with the x-axis at point $A$, and its axis of symmetry is the line $x=2$. Find the analytical expression of the parabola.

Solution:
Solution: Let $x=0$, we obtain $y=-x+3=3$,\\
Let $y=0$, we obtain $-x+3=0$,\\
Solving yields $x=3$,\\
$\therefore B(3, 0)$ and $C(0, 3)$,\\
$\because$ The axis of symmetry of the parabola is the line $x=2$, and the other intersection point of the parabola $y=ax^2+bx+c$ with the x-axis is A,\\
$\therefore A(1, 0)$,\\
Suppose the equation of the parabola is $y=a(x-1)(x-3)$,\\
Substituting $C(0, 3)$ into it, we get $3a=3$,\\
$\therefore a=1$,\\
$\therefore$ The equation of the parabola is $y=x^2-4x+3$; hence, the equation of the parabola is $\boxed{y=x^2-4x+3}$.


Question: As shown in the figure, the line $y_1=-x+3$ intersects the x-axis and the y-axis at points $B$ and $C$ respectively, and the parabola $y_2=ax^2+bx+c$ that passes through points $B$ and $C$ has another intersection point with the x-axis at point $A$, with its axis of symmetry being the line $x=2$. When $y_1<y_2$, write down directly the range of values for $x$.

Solution:
\textbf{Solution:} Solve: $x<0$ or $x>3$.

Directly write the range of $x$ as \boxed{x<0} or \boxed{x>3}.


Question:
Synthesis and Exploration\\
As shown in Figure 1, the parabola $y=-\frac{1}{2}x^{2}+\frac{3}{2}x+2$ intersects the x-axis at points A and B (where point A is to the left of point B), and intersects the y-axis at point C. Point D is on the parabola in the first quadrant. Please write down the coordinates of points A, B, and C directly.

Solution: \textbf{Solution:} Let \boxed{A(-1, 0)}, \boxed{B(4, 0)}, \boxed{C(0, 2)}


Question: Synthesis and Exploration\\ As shown in Figure 1, the parabola $y=-\frac{1}{2}x^{2}+\frac{3}{2}x+2$ intersects the x-axis at points A and B (point A is to the left of point B) and intersects the y-axis at point C. Point D lies on the parabola in the first quadrant.\\ If $ S_{\triangle COD}=\frac{1}{3}S_{\triangle OBD}$, find the coordinates of point D.

Solution:
\textbf{Solution:} Let $DG \perp x$-axis and $DH \perp y$-axis through the point D, with the feet of the perpendiculars being G and H, respectively.

Suppose $D\left(m, -\frac{1}{2}m^{2}+\frac{3}{2}m+2\right)$.

Since the area of $\triangle DCO = \frac{1}{2}OC \cdot DH = \frac{1}{2} \times 2 \times m$.

The area of $\triangle BOD = \frac{1}{2}OB \cdot DG = \frac{1}{2} \times 4 \times \left(-\frac{1}{2}m^{2}+\frac{3}{2}m+2\right)$

Since the area of $\triangle COD = \frac{1}{3}$ the area of $\triangle CBD$.

Therefore, $\frac{1}{2} \times 2 \times m = \frac{1}{3} \times \frac{1}{2} \times 4 \times \left(-\frac{1}{2}m^{2}+\frac{3}{2}m+2\right)$

Solving this, we get $m_{1}=2$, $m_{2}=-2$ (discard).

When $m=2$, $-\frac{1}{2}m^{2}+\frac{3}{2}m+2=3$.

Therefore, $D\left(2, 3\right)$.

Hence, the coordinates of point D are $\boxed{D\left(2, 3\right)}$.


Question: As shown in the figure, in the circle $\odot O$, $PA$ is a diameter, $PC$ is a chord, $PH$ bisects $\angle APB$ and intersects $\odot O$ at point $H$. A line is drawn through $H$ such that $HB\perp PC$ and it intersects the extension of $PC$ at point $B$. $HB$ is a tangent to the circle $\odot O$; if $HB=4$ and $BC=2$, what is the diameter of $\odot O$?

Solution:
\textbf{Solution:} As shown in the figure, draw OE$\perp$PC with E being the foot of the perpendicular from O, \\
since OE$\perp$PC, OH$\perp$BH, BP$\perp$BH, \\
thus, quadrilateral EOHB is a rectangle, \\
therefore, OE=BH=4, OH=BE, \\
therefore, CE=OH-2, \\
since OE$\perp$PC, \\
therefore, PE=EC=OH-2=OP-2, \\
in $\triangle POE$, $OP^{2}=PE^{2}+OE^{2}$, \\
therefore, $OP^{2}=(OP-2)^{2}+16$, \\
therefore, OP=5, \\
therefore, AP=2OP=\boxed{10}, \\
therefore, the diameter of $\odot O$ is \boxed{10}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, the angle bisector of $\angle BAC$ intersects $BC$ at point D. Point O is on $AB$, and the circle with point O as the center and $OA$ as the radius just passes through point D and intersects $AC$, $AB$ at points E, F respectively. If $BD=2\sqrt{5}$ and $BF=2$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} In the right-angled triangle $\triangle OBD$, let the radius be $r$. According to the Pythagorean theorem, we have
\[r^{2}+BD^{2}=(r+BF)^{2},\]
\[\because BD=2\sqrt{5}, BF=2,\]
\[\therefore r^{2}+(2\sqrt{5})^{2}=(r+2)^{2},\]
solving this gives $r=\boxed{4}$.


Question: As shown in the figure, point \(A\) lies on the extension line of diameter \(CD\) of circle \(\odot O\), and point \(B\) is on circle \(\odot O\), with \(\angle A=\angle C=30^\circ\). Line \(AB\) is a tangent to circle \(\odot O\); if \(BC=2\sqrt{3}\), what is the length of \(AC\)?

Solution:
\textbf{Solution:} Construct $BE\perp AC$ at point $E$\\
Since $\angle C=30^\circ$ and $BC=2\sqrt{3}$\\
It follows that $BE=\sqrt{3}$\\
Thus, $CE=\sqrt{BC^{2}-BE^{2}}=\sqrt{(2\sqrt{3})^{2}-(\sqrt{3})^{2}}=3$\\
Since $\angle A=\angle C=30^\circ$\\
It implies $AB=BC$\\
Since $BE\perp AC$\\
It concludes $AC=2EC=\boxed{6}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, point $D$ is on the extension of $AB$, and $C$ is a point on $\odot O$. Given that $\angle A=25^\circ$ and $\angle D=40^\circ$, what is the degree measure of $\angle DOC$?

Solution:
\textbf{Solution:} Given $OA = OC$, $\angle A = 25^\circ$, \\
$\therefore \angle A = \angle OCA = 25^\circ$, \\
$\therefore \angle DOC = 2\angle A = \boxed{50^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $BC$ is a chord of $\odot O$, $AE \perp OC$ at point $D$, intersecting $BC$ at $F$, and intersecting the line passing through point $B$ at point $E$, with $BE=EF$. $BE$ is a tangent line of $\odot O$; if the radius of $\odot O$ is 10, and $OD=6$, find the length of $BE$.

Solution:
\textbf{Solution}: Since the radius of circle $O$ is 10,\\
hence $OA=OB=OC=10$,\\
therefore $AB=20$,\\
since $AE\perp OC$,\\
therefore $\angle ADO=90^\circ$,\\
thus in right triangle $ADO$, $AD=\sqrt{AO^{2}-DO^{2}}$,\\
since $OD=6$,\\
therefore $AD=\sqrt{AO^{2}-DO^{2}}=\sqrt{10^{2}-6^{2}}=8$,\\
since combining (1), it is known that $\angle ABE=\angle ADO=90^\circ$ and $\angle BAE=\angle DAO$,\\
therefore $\triangle ADO\sim \triangle ABE$,\\
therefore $\frac{BE}{AB}=\frac{DO}{AD}$, that is $BE=\frac{DO}{AD}\times AB$,\\
since $AD=8$, $AB=20$, $DO=6$,\\
therefore $BE=\frac{DO}{AD}\times AB=\frac{6}{8}\times 20=\boxed{15}$,\\
that is, the sought value is \boxed{15}.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and point $P$ is on $\odot O$ with $PA=PB$. Point $M$ is outside $\odot O$, and $MB$ is tangent to $\odot O$ at point $B$. Connect $OM$, and draw $AC\parallel OM$ through point $A$ to intersect with $\odot O$ at point $C$, and connect $BC$ to intersect $OM$ at point $D$. $MC$ is a tangent to $\odot O$. Given that $AB=20$ and $BC=16$, connect $PC$. What is the length of $PC$?

Solution:
\textbf{Solution:} Since $AB$ is the diameter of $\odot O$,\\
it follows that $\angle ACB=\angle APB=90^\circ$,\\
Since $AB=20$,\\
we have $PA=PB=\frac{\sqrt{2}}{2}\times 20=10\sqrt{2}$,\\
Since $BC=16$,\\
we get $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{20^{2}-16^{2}}=12$,\\
Draw $AH\perp PC$ at point H through point A,\\
thus $\angle AHC=\angle AHP=90^\circ$,\\
Since $\angle ACH=\angle ABP=45^\circ$,\\
we have $AH=CH=AC\times \sin 45^\circ=12\times \frac{\sqrt{2}}{2}=6\sqrt{2}$,\\
Therefore, $PH=\sqrt{PA^{2}-AH^{2}}=\sqrt{(10\sqrt{2})^{2}-(6\sqrt{2})^{2}}=8\sqrt{2}$,\\
Hence, $PC=PH+CH=\boxed{14\sqrt{2}}$.


Question: As shown in the figure, $AC$ is the diameter of the circle $\odot O$, $P$ is the midpoint of the semicircle $AC$. Line $AP$ is extended to point $B$ such that $PB=AP$. Lines $CP$, $CB$, and $OP$ are drawn. $BC$ is a tangent of circle $\odot O$. If $AC=4$, what is the area of the shaded section in the figure?

Solution:
\textbf{Solution:} Given $AC=4$,\\
thus $BC=AC=4$, $OA=OP=2$,\\
since $P$ is the midpoint of the semicircle $AC$,\\
hence $AC\perp OP$,\\
therefore $S_{\text{shaded}}=S_{\triangle ABC}-S_{\triangle AOP}-S_{\text{sector} POC}=\frac{1}{2}\times 4\times 4-\frac{1}{2}\times 2\times 2-\frac{90^\circ\pi \times 2^2}{360}=\boxed{6-\pi}$


Question: As shown in the figure, the diameter AB of the circle $\odot O$ forms an angle $\angle A=30^\circ$ with the chord AC, where $AC=CP$. $CP$ is a tangent to $\odot O$; if $PC=6$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:}

Since $AC=PC$,

it follows that $\angle P=\angle A=30^\circ$,

thus $\angle COB=60^\circ$ and $OP=2OC$.

Given $PC=6$,

we have $OP^2-OC^2=6^2$,

which leads to $OC=2\sqrt{3}$.

Therefore, the area of the shaded region in the diagram $S_{\triangle POC}-S_{\text{sector} COB}$

$=\frac{1}{2}\times 6\times 2\sqrt{3}-\frac{60^\circ \cdot \pi \times (2\sqrt{3})^2}{360^\circ}=6\sqrt{3}-2\pi$.

Hence, the area of the shaded region in the diagram is $\boxed{6\sqrt{3}-2\pi}$.


Question: As shown in the diagram, points $A,B$ lie on circle $O$, and the bisector of $\angle BAO$ intersects circle $O$ at point $D$. Point $C$ is on the extension of $OA$ such that $\angle CBA=\angle D$. $CB$ is a tangent to circle $O$; If $DB\parallel OA$ and $BD=3$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} Method 1:\\
Since $DB\parallel OA$,\\
it follows that $\angle OAD=\angle D$,\\
Since $AD$ is the bisector of $\angle BAO$,\\
it follows that $\angle OAD=\angle BAD$,\\
Therefore $\angle BAD=\angle D$, hence $BD=AB=3$,\\
Since $\angle AOB=2\angle E=2\angle D=2\angle BAD=\angle OAB$, and $OA=OB$,\\
it follows that $\triangle AOB$ is an equilateral triangle, hence $OA=AB$,\\
Therefore, $OA=\boxed{3}$, which means the radius of circle O is 3.\\
Method 2:\\
Connect $DO$, where $OD=OA$,\\
Therefore $\angle ODA=\angle OAD$,\\
Moreover, since $\angle OAD=\angle BAD$,\\
Therefore $\angle ODA=\angle BAD$,\\
Therefore $OD\parallel AB$ (since the alternate interior angles are equal, the lines are parallel),\\
Also, since $DB\parallel OA$,\\
Therefore the quadrilateral $OABD$ is a parallelogram,\\
Hence, $OA=BD=\boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, we have $AB=AC$. A circle $\odot O$ with diameter $AC$ intersects $BC$ at point $D$. A line through point $D$ and perpendicular to $AB$ is drawn, with the foot of the perpendicular being point $E$. Extend $BA$ to intersect $\odot O$ at point $F$. $DE$ is a tangent to $\odot O$. Given $DE=2$ and $AF=3$, find the length of $AE$.

Solution: Given that the provided information is only the number "1", there is not enough context or specific LaTeX code to perform any cleaning or modification. Please provide the complete LaTeX code or a more detailed description of the problem, so that I can offer accurate assistance.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and lines $BD$ and $CD$ are the tangents to $\odot O$ at points $B$ and $C$, respectively. If $BD=2$, what is the length of $CD$?

Solution: \textbf{Solution:} Due to insufficient information provided in the question, it is impossible to offer a \boxed{solution}.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and the lines $BD$ and $CD$ are tangents to $\odot O$ at points $B$ and $C$, respectively. If $\angle BDC = 130^\circ$, what is the degree measure of $\angle A$?

Solution:
\textbf{Solution:} Let's connect $OC$, $BC$.\\
Since $BD$, $CD$ are tangents to the circle centered at $O$ through points $B$ and $C$ respectively,\\
Therefore $OC\perp CD$, $OB\perp BD$,\\
Therefore $\angle OCD=\angle OBD=90^\circ$,\\
Since $\angle BDC=130^\circ$,\\
Therefore $\angle BOC=360^\circ-\angle OCD-\angle BDC-\angle OBD=50^\circ$,\\
Therefore $\angle A=\frac{1}{2}\angle BOC=\boxed{25^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, with $C$ and $D$ being two points on $\odot O$, $\angle BAC = \angle DAC$. A line $EF \perp AD$ is drawn through point $C$, intersecting the extension of $AD$ at point $E$, and $BC$ is connected. $EF$ is the tangent of $\odot O$; if $DE=2$ and $BC=4$, what is the length $l$ of the minor arc $\overset{\frown}{BC}$?

Solution:
\textbf{Solution:} Draw $OD$, $DC$,\\
since $\angle DAC = \frac{1}{2}\angle DOC$, $\angle OAC = \frac{1}{2}\angle BOC$, $\angle DAC = \angle OCA = \angle OAC$,\\
thus $\angle DOC = \angle BOC$, $DC=BC=4$,\\
therefore, in $\triangle CED$, $\sin\angle ECD = \frac{DE}{DC} = \frac{1}{2}$,\\
thus $\angle ECD = 30^\circ$,\\
also, since $EF$ is a tangent to $\odot O$,\\
thus $\angle OCD = 60^\circ$,\\
therefore, $\triangle COD$ is an equilateral triangle,\\
thus $\angle BOC = \angle COD = 60^\circ$,\\
therefore, $\triangle BOC$ is an equilateral triangle,\\
thus $OC = BC = 4$,\\
therefore, $l = \frac{60\pi \times 4}{180} = \boxed{\frac{4}{3}\pi}$.


Question: As shown in the figure, AB is the diameter of the circle $\odot$O, and points F, C are on circle $\odot$O, with $\overset{\frown}{AF}=\overset{\frown}{FC}=\overset{\frown}{CB}$. Connect AC and AF, and draw $CD\perp AF$ extending through point C to meet the extension of AF at point D, with D being the foot of the perpendicular. CD is the tangent of circle $\odot$O; If CD=$2\sqrt{3}$, what is the radius of $\odot$O?

Solution:
\textbf{Solution:} Connect BC as shown in the figure, \\
since AB is the diameter, \\
therefore, $\angle ACB=90^\circ$, \\
since $\overset{\frown}{AF}=\overset{\frown}{FC}=\overset{\frown}{CB}$, \\
therefore, $\angle BOC= \frac{1}{3}\times180^\circ=60^\circ$, \\
therefore, $\angle BAC=30^\circ$, \\
therefore, $\angle DAC=30^\circ$, \\
in $\triangle ADC$, where CD=$2\sqrt{3}$, \\
therefore, AC=2CD=$4\sqrt{3}$, \\
in $\triangle ACB$, where BC$^{2}$+AC$^{2}$=AB$^{2}$, \\
that is, ($4\sqrt{3}$)$^{2}$+($\frac{1}{2}$AB)$^{2}$=AB$^{2}$, \\
therefore, AB=8, \\
therefore, the radius of $\odot O$ is $\boxed{4}$.


Question: As shown in the figure, $AD$ is tangent to the circle $\odot O$ at point $D$, and point $A$ is on the extension of the diameter $CB$. $\angle DCB = \angle ADB$; If $\angle DCB = 30^\circ$ and $AC = 3\sqrt{3}$, find the length of $AD$.

Solution:
\textbf{Solution:} Since $\because \angle DCB = \angle ADB$ and $\angle DAC = \angle CAD$,\\
$\therefore \triangle ADB \sim \triangle ACD$,\\
$\therefore \frac{DB}{CD} = \frac{AD}{AC}$,\\
Since $CB$ is the diameter,\\
$\therefore \angle CDB = 90^\circ$, $\angle DCB = 30^\circ$,\\
$\therefore \tan \angle DCB = \frac{DB}{CD} = \frac{\sqrt{3}}{3}$,\\
$\therefore \frac{AD}{AC} = \frac{\sqrt{3}}{3}$,\\
Since $AC = 3\sqrt{3}$,\\
$\therefore AD = \boxed{3}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$ and $\angle B=30^\circ$. Point $D$ is the midpoint of side $AB$, and point $O$ is on side $BC$. A circle with center $O$ passes through vertex $C$ and intersects side $AB$ at point $D$. The line $AB$ is a tangent to circle $O$; if $AC=\sqrt{3}$, what is the area of the shaded region?

Solution:
\textbf{Solution:} From (1) we know: $AC=AD=BD=\frac{1}{2}AB$,\\
and since $AC=\sqrt{3}$,\\
therefore $BD=AC=\sqrt{3}$,\\
since $\angle B=30^\circ$, $\angle BDO=\angle ADO=90^\circ$,\\
therefore $\angle BOD=60^\circ$, $BO=2DO$,\\
by the Pythagorean theorem we have: $BO^2=OD^2+BD^2$,\\
which gives $(2OD)^2=OD^2+(\sqrt{3})^2$,\\
solving this yields: $OD=1$ (negatives disregarded),\\
thus the area of the shaded region $S=S_{\triangle BDO}-S_{\text{sector} DOE}=\frac{1}{2}\times 1\times \sqrt{3}-\frac{60^\circ\pi \times 1^2}{360}=\boxed{\frac{\sqrt{3}}{2}-\frac{\pi}{6}}$.


Question: As shown in the diagram, $AB$ is the diameter of $\odot O$, the chord $AC = BC$, and $E$ is the midpoint of $OB$. Extend line $CE$ to point $F$ such that $EF=CE$, and draw line $AF$ intersecting $\odot O$ at point $D$. Draw lines $BD$ and $BF$. The line $BF$ is tangent to $\odot O$; if the length of $AF$ is $5\sqrt{2}$, find the radius of $\odot O$ and the length of $BD$.

Solution:
\textbf{Solution:} Let the radius of $\odot O$ be $r$, hence $AB=2r$, $BF=OC=r$. In $\triangle ABF$, we have: ${\left(2r\right)}^{2}+{r}^{2}={\left(5\sqrt{2}\right)}^{2}$.

$\therefore$ We only take $r=\sqrt{10}$, thus the radius of $\odot O$ is $\boxed{\sqrt{10}}$.

$\because$ $AB$ is the diameter of $\odot O$, thus $AB=2\sqrt{10}$,

$\therefore$ $BD\perp AF$,

$\therefore$ $BF=\sqrt{AF^{2}-AB^{2}}=\sqrt{{\left(5\sqrt{2}\right)}^{2}-{\left(2\sqrt{10}\right)}^{2}}=\sqrt{10}$,

$\because$ AB is a diameter,

$\therefore$ $\angle ADB=90^\circ$,

$\therefore$ ${s}_{\triangle ABF}=\frac{1}{2}AB\times BF=\frac{1}{2}AF\times BD$,

$\therefore$ $\frac{1}{2}\times 2\sqrt{10}\times \sqrt{10}=\frac{1}{2}\times 5\sqrt{2}\times BD$, solving this yields $BD=\boxed{2\sqrt{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, point D and E are on $\odot O$, and quadrilateral BDEO is a parallelogram. A line through point D, $DC\perp AE$, intersects the extension of AE at point C. CD is the tangent of $\odot O$. If $AC=9$, what is the area of the shaded region?

Solution:
\textbf{Solution:}

Given (1) we have $\angle EAO = \angle AOE = \angle OBD = \angle BOD = 60^\circ$, and $ED \parallel AB$,

$\therefore \angle EAO = \angle CED = 60^\circ$,

$\because \angle AOE + \angle EOD + \angle BOD = 180^\circ$,

$\therefore \angle EOD = 60^\circ$,

$\therefore \triangle DEO$ is an equilateral triangle,

$\therefore ED = OE = AE$,

$\because CD \perp AE$, $\angle CED = 60^\circ$,

$\therefore \angle CDE = 30^\circ$,

$\therefore ED = 2CE = AE$,

$\because AC = 9$,

$\therefore CE = 3$, $AE = OE = ED = 6$,

$\therefore CD = \sqrt{ED^2 - CE^2} = 3\sqrt{3}$,

Let the height of $\triangle OED$ be $h$,

$\therefore h = OE \cdot \sin 60^\circ = 3\sqrt{3}$,

$\therefore S_{\text{sector} ED} = S_{\text{sector} OED} - S_{\triangle OED} = \frac{\pi r^{2}n}{360}-\frac{1}{2}ED\cdot h = 6\pi - 9\sqrt{3}$,

$\therefore S_{\text{shaded area}} = S_{\triangle CED} - S_{\text{arc} ED} = \frac{1}{2}CE \cdot CD - (6\pi - 9\sqrt{3}) = \boxed{\frac{27\sqrt{3}}{2} - 6\pi}$.


Question: As shown in the figure, in the right-angled triangle $\triangle ABC$ with $\angle C=90^\circ$, point D is on AB. The circle $O$ with diameter AD intersects BC at point E, and AC at point F, with AE bisecting $\angle BAC$. BC is a tangent to circle $O$. Given that $\angle EAB=30^\circ$ and $OD=5$, find the perimeter of the shaded area in the diagram.

Solution:
\textbf{Solution:} Given that $\because \angle EAB=30^\circ$, \\
$\therefore \angle EOD=60^\circ$, \\
$\therefore \angle OEB=90^\circ$, \\
$\therefore \angle B=30^\circ$, \\
$\therefore OB=2OE=2OD=10$, \\
$\therefore BD=5$, \\
$\therefore BE= \sqrt{OB^{2}-OE^{2}} = \sqrt{10^{2}-5^{2}}=5\sqrt{3}$, \\
$\therefore$ the length of arc DE is $= \frac{60^\circ \pi \times 5}{180} = \frac{5\pi }{3}$, \\
$\therefore C_{\text{shaded}}= BD+BE+l_{\overset{\frown}{DE}} = 5+5\sqrt{3}+\frac{5\pi}{3} = \boxed{5+5\sqrt{3}+\frac{5\pi}{3}}$.


Question: Given, as shown in the figure, line MN intersects $\odot O$ at points A and B, AC is a diameter, DE is tangent to $\odot O$ at point D, and DE is drawn perpendicular to MN at point E. AD bisects $\angle CAE$; If AE＝2 and AD＝4, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Draw OF$\perp$AB through point O, it's clear that the quadrilateral ODEF is a rectangle,\\
hence, OF=DE, OD=EF,\\
Let the radius of the circle OD=EF=OA=$r$,\\
since AE=2, AD=4, and $\angle$AED=90$^\circ$,\\
therefore, DE=$\sqrt{AD^2-AE^2}=2\sqrt{3}$,\\
thus, OF=DE=$2\sqrt{3}$, AF=EF−AE=$r−2$,\\
In $\triangle$AOF, according to the Pythagorean theorem, we have: OA$^2$=AF$^2$+OF$^2$, i.e., $r^2$=$(r−2)^2+(2\sqrt{3})^2$,\\
Solving this, we find: $r=\boxed{4}$,\\
Hence, the radius of circle O is \boxed{4}.


Question: As shown in the figure, $\triangle ABC$ is an inscribed triangle of circle $\odot O$ with $AB$ as the diameter. $BD$ is tangent to $\odot O$ at point $B$, and intersects the extension line of $AC$ at point $D$. $E$ is the midpoint of $BD$, and $CE$ intersects the extension line of $BA$ at point $F$. Given that $BD=8$, and $BE = \frac{1}{3}EF$. $FC$ is a tangent line of $\odot O$; find the length of $AF$?

Solution:
\textbf{Solution:} Given $BD=8$ and point $E$ is the midpoint of $BD$,\\
hence $BE= \frac{1}{2}BD=4$,\\
since $BE= \frac{1}{3}EF$,\\
thus $EF=3BE=12$,\\
in $\triangle FBE$, $BF= \sqrt{EF^{2}-BE^{2}}=\sqrt{12^{2}-4^{2}}=8\sqrt{2}$,\\
from (1) we have $\angle OCF=\angle ABD=90^\circ$,\\
since $\angle F=\angle F$,\\
thus $\triangle FOC\sim \triangle FEB$,\\
hence $\frac{OC}{BE}=\frac{OF}{EF}$,\\
let $OC=x$, then $OF=BF-OB=8\sqrt{2}-x$,\\
hence $\frac{x}{4}=\frac{8\sqrt{2}-x}{12}$,\\
thus $x=2\sqrt{2}$,\\
hence $AF=8\sqrt{2}-2x=4\sqrt{2}$;\\
Therefore, the length of $AF$ is $\boxed{4\sqrt{2}}$.


Question: As shown in the diagram, $\triangle ABC$ is an inscribed triangle of circle $\odot O$ with $AB$ as the diameter. $BD$ is tangent to $\odot O$ at point $B$ and intersects the extension line of $AC$ at point $D$. $E$ is the midpoint of $BD$, and $CE$ intersects the extension line of $BA$ at point $F$. $BD=8$, and $BE = \frac{1}{3}EF$. If $\angle F= 20^\circ$ and $BC=3\sqrt{2}$, what is the area of the shaded region in the diagram?

Solution:
\textbf{Solution:} Let OM $\perp$ BC at point M,\\
$\therefore$ BM=$\frac{1}{2}$BC=$\frac{3\sqrt{2}}{2}$,\\
In $\triangle BMO$, OM=$\sqrt{OB^{2}-BM^{2}}=\sqrt{(2\sqrt{2})^{2}-\left(\frac{3\sqrt{2}}{2}\right)^{2}}=\frac{\sqrt{14}}{2}$,\\
$\therefore$ Area of $\triangle BOC$=$\frac{1}{2}$BC$\cdot$OM=$\frac{1}{2}\times 3\sqrt{2}\times\frac{\sqrt{14}}{2}=\frac{3\sqrt{7}}{2}$,\\
$\because \angle F=20^\circ$,\\
$\therefore \angle BOC=\angle F+\angle OCF=110^\circ$,\\
$\therefore$ Area of the sector BOC=$\frac{110}{360}\pi \times (2\sqrt{2})^{2}=\frac{22}{9}\pi$,\\
$\therefore$ Area of the shaded region=Area of the sector BOC-Area of $\triangle BOC$=$\frac{22}{9}\pi -\frac{3\sqrt{7}}{2}$,\\
$\therefore$ The area of the shaded part in the figure is $\boxed{\frac{22}{9}\pi -\frac{3\sqrt{7}}{2}}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, and $OC$ is the radius. Connect $AC$ and $BC$. Extend $OC$ to point $D$ such that $\angle CAD = \angle B$. Draw $DM\perp AD$ meeting the extension of $AC$ at point $M$. $AD$ is the tangent line of $\odot O$. If $AD=6$ and $\tan\angle CAD=\frac{1}{2}$, find the length of the radius of $\odot O$.

Solution:
\textbf{Solution:}\\
Because $\tan\angle CAD=\frac{1}{2}=\frac{DM}{AD}$, and $AD=6$,\\
then $DM=3$,\\
Since $OA=OC$,\\
it follows that $\angle OAC=\angle OCA$,\\
Given $AD\perp OA$ and $DM\perp AD$,\\
it implies $OA\parallel DM$,\\
thus $\angle M=\angle OAC$,\\
Since $\angle OCA=\angle DCM$,\\
it results in $\angle DCM=\angle M$,\\
therefore $DC=DM=3$,\\
In $\triangle OAD$, using $OA^2+AD^2=OD^2$, which means $OA^2+6^2=(OC+3)^2=(OA+3)^2$,\\
thus $OA=\frac{9}{2}$,\\
Hence, the radius of $\odot O$ is $\boxed{\frac{9}{2}}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$. Through point $O$, draw $OC\perp OA$, where $OC$ intersects $AB$ at $P$, and $CP=BC$. Point $Q$ lies on the arc $\overset{\frown}{AmB}$. $BC$ is a tangent line to the circle $\odot O$; Given that $\angle BAO=25^\circ$, what is the measure of $\angle AQB$?

Solution:
\textbf{Solution:} Since $\angle BAO=25^\circ$,\\
it follows that $\angle ABO=25^\circ$, $\angle APO=65^\circ$,\\
thus $\angle POB=\angle APO-\angle ABO=40^\circ$,\\
therefore $\angle AQB=\frac{1}{2}(\angle AOP+\angle POB)=\frac{1}{2}\times 130^\circ=\boxed{65^\circ}$.


Question: As shown in the figure, $AB$ is the diameter of $\odot O$. Points $F$ and $C$ are on $\odot O$, and arc $AF = \text{arc} FC = \text{arc} BC$. Connect $AC$ and $AF$. Through point $C$, draw $CD \perp AF$ to intersect the extended line of $AF$ at point $D$, with the foot of the perpendicular being $D$. $CD$ is a tangent of $\odot O$; If $CD = 2\sqrt{3}$, what is the radius of $\odot O$?

Solution:
\textbf{Solution:} Connect BC as shown in the figure. \\
Since AB is the diameter, it follows that $\angle ACB=90^\circ$. Since $\overset{\frown}{AF} = \overset{\frown}{FC} = \overset{\frown}{BC}$, it follows that $\angle BOC = \frac{1}{3} \times 180^\circ = 60^\circ$.\\
Therefore, $\angle BAC = 30^\circ$, thus $\angle DAC = 30^\circ$. In $\triangle ADC$, given $CD = 2\sqrt{3}$, it follows that $AC = 2CD = 4\sqrt{3}$.\\
In $\triangle ACB$, $BC = \frac{\sqrt{3}}{3}AC = \frac{\sqrt{3}}{3} \times 4\sqrt{3} = 4$, therefore $AB = 2BC = 8$.\\
Hence, the radius of the circle $O$ is $\boxed{4}$.


Question: As illustrated in the diagram, in $\triangle ABC$, $\angle ACB = 90^\circ$, $AC = BC$, and point $O$ lies inside $\triangle ABC$. Circle $O$ passes through points $B$ and $C$, and intersects line $AB$ at point $D$. Extend line $CO$ to intersect line segment $AB$ at point $G$, and construct parallelogram $GDEC$ with $GD$ and $GC$ as adjacent sides. Line $DE$ is a tangent to circle $O$; given $DE = 7$ and $CE = 5$, what is the radius of circle $O$?

Solution:
\textbf{Solution:} Let the radius of $\odot O$ be $r$,\\
Since quadrilateral GDEC is a parallelogram,\\
it follows that CG = DE = 7, and DG = CE = 5,\\
Since $\angle GOD=90^\circ$,\\
it follows that OD$^{2}$+OG$^{2}$=DG$^{2}$, that is, $r^{2}+(7-r)^{2}=5^{2}$,\\
Solving this gives: $r_{1}=3$, $r_{2}=4$,\\
When $r=3$, OG = 4 > 3, meaning point G is outside $\odot O$, which is against the problem statement, thus discarded,\\
Therefore, $r=\boxed{4}$, meaning the radius of $\odot O$ is \boxed{4}.


Question: As shown in the diagram, $\odot O$ is the circumcircle of the right-angled triangle $ABC$, with the diameter $AC=4$. A tangent line is drawn through point $C$ to $\odot O$, intersecting the extended line of $AB$ at point $D$. $M$ is the midpoint of $CD$, and lines $BM$ and $OM$ are drawn, with $BC$ and $OM$ intersecting at point $N$. $BM$ is tangent to $\odot O$; when $\angle BAC=60^\circ$, what is the area of the shape enclosed by chord $AB$ and arc $AB$?

Solution:
\textbf{Solution:} Since $\angle BAC=60^\circ$ and $OA=OB$, \\
it follows that $\triangle ABO$ is an equilateral triangle, \\
which implies that $\angle AOB=60^\circ$.\\
Given that $AC=4$, it follows that $OA=2$,\\
therefore, the area of the region bounded by chord $AB$ and arc $AB$ $=S_{\text{sector}}AOB-S_{\triangle}AOB$\\
$=\frac{60^\circ\pi \times 2^2}{360^\circ}-\frac{\sqrt{3}}{4}\times 2^2=\boxed{\frac{2\pi}{3}-\sqrt{3}}$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, $PA$ and $PC$ are tangents to $\odot O$, with $A$ and $C$ being the points of tangency. Connect $AC$ and $PO$, intersecting at point $D$. $\angle BAC=\angle OPC$; extend $PO$ to intersect $\odot O$ at point $E$, and connect $BE$ and $CE$. If $\angle BEC=30^\circ$ and $PA=8$, what is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the diagram, since $\angle BEC=30^\circ,$\\
thus $\angle BAC=30^\circ,$ and $AC\perp OP,$\\
thus $\angle AOP=60^\circ,$\\
since $\angle OAP=90^\circ, PA=8,$\\
thus $\tan\angle AOP=\frac{PA}{AO}=\tan60^\circ=\sqrt{3},$\\
thus $AO=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3},$\\
thus $AB=2AO=\boxed{\frac{16\sqrt{3}}{3}}.$


Question: As shown in the figure, $AB$ is the diameter of $\odot O$, the chord $CD\perp AB$ at point $E$, and chord $AF$ intersects chord $CD$ at point $G$, with $AG=CG$. A perpendicular line to $BF$ through point $C$ intersects the extension of $BF$ at point $H$. $CH$ is the tangent of $\odot O$; if $FH=2$ and $BF=4$, what is the length of arc $CD$?

Solution:
\textbf{Solution}: From (1) we have: $BH \perp CH$, $OC \perp CH$,\\
$\therefore OC \parallel BH$,\\
$\because CH \parallel AF$,\\
$\therefore$ Quadrilateral $CMFH$ is a parallelogram,\\
$\because OC \perp CH$,\\
$\therefore \angle OCH = 90^\circ$,\\
$\therefore$ Quadrilateral $CMFH$ is a rectangle,\\
$\therefore OC \perp AF$, $CM = HF = 2$,\\
$\therefore AM = FM$,\\
$\because$ Point $O$ is the midpoint of $AB$,\\
$\therefore OM = \frac{1}{2} BF = 2$,\\
$\therefore CM = OM$,\\
$\therefore OC = 4$, $AM$ perpendicularly bisects $OC$,\\
$\therefore AC = AO$,\\
and $AO = OC$,\\
$\therefore AC = OC = OA$,\\
$\therefore \triangle AOC$ is an equilateral triangle,\\
$\therefore \angle AOC = 60^\circ$,\\
$\because \overset{\frown}{AC} = \overset{\frown}{AD}$,\\
$\therefore \angle AOD = \angle AOC = 60^\circ$,\\
$\therefore \angle COD = 120^\circ$,\\
$\therefore$ The length of arc $CD$ is $\frac{120 \times \pi \times 4}{180} = \boxed{\frac{8}{3}\pi}$.


Question: As shown in the diagram, in $ \triangle ABC$, with $ \angle C=90^\circ$, AC and BC are tangent to circle $ \odot O$ at points E and F, respectively. AB passes through point M on circle $ \odot O$, and $ AE=AM$. AB is a tangent to circle $ \odot O$; if $ AC=8$ and $ BC=6$, what is the radius of circle $ \odot O$?

Solution:
\textbf{Solution:} Connect OF. Let the radius of $\odot O$ be $r$.\\
Since BC is tangent to $\odot O$ at point F,\\
thus OF$\perp$BC,\\
therefore $\angle$OFC$=90^\circ$,\\
Also, since $\angle$C$=90^\circ$, $\angle$OEC$=90^\circ$, and OF=OE,\\
therefore, quadrilateral OFCE is a square,\\
thus CF=CE=OE$=r$,\\
Since AB, BC, AC are all tangent to $\odot O$,\\
therefore BM=BF$=6-r$, AM=AE$=8-r$,\\
In $\triangle ABC$, $AB=\sqrt{BC^2+AC^2}=10$,\\
Since BM+AM=AB,\\
therefore $6-r+8-r=10$,\\
therefore $r=\boxed{2}$\\
thus the radius of $\odot O$ is \boxed{2}.


Question: As shown in the diagram, it is known that $AB$ is the diameter of $\odot O$, and $C$ is a point on $\odot O$. The bisector of $\angle OCB$ intersects $\odot O$ at point $D$. A tangent to $\odot O$ at point $D$ intersects the extension of line $CB$ at point $E$. $CE \perp DE$; if $AB=10$, and $\tan A= \frac{1}{3}$, find the length of $DE$.

Solution:
\textbf{Solution:} Given $OD\parallel CE$,\\
it follows that $\angle ODC=\angle DCE$.\\
Given $OD=OA$ and $OC=OD$,\\
it implies $\angle A=\angle ADO$ and $\angle ODC=\angle OCD$.\\
Since $\angle A=\angle DCE$,\\
we have $\angle A=\angle OCD=\angle ADO=\angle CDO$.\\
Given $OA=OC$,\\
it results in $\triangle ADO\cong \triangle CDO$ (AAS),\\
thus, $CD=AD$.\\
Since $AB$ is the diameter, we have $\angle ADB=90^\circ$,\\
and given $\tan A=\frac{DB}{AD}=\frac{1}{3}$ and $AB=10$,\\
from $AD^2+DB^2=AB^2=100$\\
we find $BD=\sqrt{10}$, $AD=3\sqrt{10}$,\\
thus, $CD=AD=3\sqrt{10}$.\\
Since $\angle A=\angle DCE$ and $\angle ADB=\angle E=90^\circ$,\\
$\triangle ADB\sim \triangle CED$,\\
it follows that $\frac{AB}{CD}=\frac{BD}{DE}$,\\
thus, $\frac{10}{3\sqrt{10}}=\frac{\sqrt{10}}{DE}$,\\
resulting in $DE=\boxed{3}$.


Question: As shown in the figure, within circle $\odot O$, point C is on the extension line of diameter AB, point D is a point on the circle above diameter AB, and CD is connected such that $\angle A = \angle BDC$. CD is a tangent of $\odot O$. If CE bisects $\angle ACD$ and intersects AD, BD at points E, F, respectively, when $DE = 2$, what is the length of $EF$?

Solution:
\textbf{Solution:} Since $CE$ bisects $\angle ACD$,\\
it follows that $\angle DCE = \angle ACE$.\\
Additionally, since $\angle A=\angle BDC$,\\
it implies that $\angle A+\angle ACE=\angle BDC+\angle DCE$,\\
which means that $\angle DEF=\angle DFE$.\\
Given that $\angle ADB=90^\circ$ and $DE=2$,\\
it follows that $DF=DE=2$,\\
hence $EF=\sqrt{DE^2+DF^2}=2\sqrt{2}=\boxed{2\sqrt{2}}$


Question: As shown in the diagram, in $\triangle ABC$, $\angle ACB=90^\circ$, point O is a point on side BC, a circle with center O and radius OB intersects side AB at point D, connect DC, and $DC=AC$. $DC$ is the tangent to $\odot O$; If the radius of $\odot O$ is 3, and $CD=4$, find the length of BC.

Solution:
Solution: In $\triangle \mathrm{ODC}$, we have $OD=3$ and $CD=4$, \\
$\therefore OC=\sqrt{OD^2+CD^2}=5$ \\
$\therefore BC=OB+OC=3+5=\boxed{8}$.


Question: As shown in the diagram, it is known that $AB$ is the diameter of $\odot O$, $AD$ is a chord of $\odot O$, $BC$ is a tangent of $\odot O$ with the point of tangency at B, and points D, F are the trisection points of the arc $\overset{\frown}{ADB}$. The extensions of $BA$ and $CD$ intersect at point E. $DC$ is a tangent of $\odot O$. If the radius of $\odot O$ is 1, calculate the area of the shaded region.

Solution:
Solution: From the construction, we have
\[
\because OD\perp EC,
\]
\[
\therefore \angle EDO=90^\circ,
\]
\[
\therefore \text{in} \triangle DOE, \angle OED + \angle DOE = 90^\circ,
\]
\[
\because \angle DOE=60^\circ,
\]
\[
\therefore \angle DEO=30^\circ,
\]
\[
\therefore \text{in} \triangle DOE, OE=2OD=2,
\]
\[
\therefore \text{in} \triangle DOE, \text{by the Pythagorean theorem}: DE=\sqrt{OE^2-OD^2}=\sqrt{2^2-1^2}=\sqrt{3},
\]
\[
\therefore S_{\triangle DOE}=\frac{1}{2}\times 1\times \sqrt{3}=\frac{\sqrt{3}}{2},
\]
\[
S_{\text{sector} DOE}=\frac{60^\circ \pi \times 1^2}{360}=\frac{\pi}{6},
\]
\[
\therefore S_{\text{shaded}}=S_{\triangle DOE}-S_{\triangle AOD}=\boxed{\frac{\sqrt{3}}{2}-\frac{\pi}{6}}.
\]


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\odot O$ is the inscribed circle of $\triangle ABC$, and D, E, F are the points of tangency. Quadrilateral ODCE is a square; if $AB=5$ and $AC=3$, what is the radius of the inscribed circle $\odot O$?

Solution:
\textbf{Solution:} Let the radius of circle $ \odot O$ be $r$,\\
since quadrilateral $ODCE$ is a square,\\
therefore $OD=DC=CE=r$,\\
in $\triangle ABC$, $BC=\sqrt{AB^{2}-AC^{2}}=4$,\\
therefore $BD=4-r$, $AE=3-r$,\\
since circle $ \odot O$ is tangent to the sides of $\triangle ABC$ at points $D$, $E$, and $F$,\\
therefore $AE=AF=3-r$, $BF=BD=4-r$,\\
and since $AB=AF+BF=5$,\\
therefore $3-r+4-r=5$, solving this gives $r=\boxed{1}$\\
therefore, the radius of the inscribed circle is \boxed{1}.


Question: Given that $AB$ is the diameter of $\odot O$, and the chord $CD$ intersects $AB$ at point $E$. A tangent to $\odot O$ at point $C$ intersects the extension of $BA$ at point $P$, with $\angle BPC=38^\circ$. Connect $OD$. If $D$ is the midpoint of $\overset{\frown}{AB}$, what is the measure of $\angle ODC$?

Solution:
\textbf{Solution:} As shown in Figure 1, connect $OC$, \\
since $D$ is the midpoint of $\overset{\frown}{AB}$,\\
it follows that $\overset{\frown}{AD}=\overset{\frown}{BD}$,\\
thus $\angle AOD=\angle BOD=\frac{1}{2}\angle AOB=90^\circ$,\\
since $PC$ is the tangent to the circle $\odot O$ at point $C$,\\
it follows that $\angle PCO=90^\circ$,\\
thus $\angle POC=90^\circ-\angle BPC=90^\circ-38^\circ=52^\circ$,\\
therefore, $\angle COD=\angle POC+\angle AOD=52^\circ+90^\circ=142^\circ$,\\
since $OC=OD$,\\
it follows that $\angle OCD=\angle ODC$,\\
thus $\angle OCD=\angle ODC=\frac{180^\circ-\angle COD}{2}=\frac{180^\circ-142^\circ}{2}=19^\circ$\\
hence, $\angle ODC=\boxed{19^\circ}$;


Question: As shown in the diagram, the diameter $AB$ of the circle $\odot O$ is $2$, point $C$ is on the circumference of the circle, and $\angle CAB=30^\circ$. Point $D$ is a moving point on the circle, $DE\parallel AB$ intersects the extension line of $CA$ at point $E$, and $CD$ is connected, intersecting $AB$ at point $F$. As shown in Figure 1, when $DE$ is tangent to $\odot O$, what is the degree measure of $\angle CFB$?

Solution:
\textbf{Solution:} As shown in the figure: Connect OD\\
$\because$ DE is tangent to $\odot$O\\
$\therefore \angle$ODE$=90^\circ$\\
$\because$ AB $\parallel$ DE\\
$\therefore \angle$AOD$+\angle$ODE$=180^\circ$\\
$\therefore \angle$AOD$=90^\circ$\\
$\because \angle$AOD$=2\angle$C\\
$\therefore \angle$C$=45^\circ$\\
$\because \angle$CFB$=\angle$CAB$+\angle$C\\
$\therefore \angle$CFB$=\boxed{75^\circ}$


Question: As shown in the figure, the diameter $AB$ of the circle $\odot O$ is $2$. Point $C$ is on the circumference, and $\angle CAB=30^\circ$. Point $D$ is a moving point on the circle, with $DE\parallel AB$ intersecting the extension of $CA$ at point $E$, and $CD$ is connected intersecting $AB$ at point $F$. As shown in figure 2, when point $F$ is the midpoint of $CD$, find the area of $\triangle CDE$.

Solution:
\textbf{Solution:} As shown in the figure: connect OC\\
$\because$ AB is the diameter, and point F is the midpoint of CD\\
$\therefore$ AB$\perp$CD, CF=DF,\\
$\because$ $\angle$COF$=2\angle$CAB$=60^\circ$,\\
$\therefore$ OF$=\frac{1}{2}$OC$=\frac{1}{2}$, CF$=\sqrt{3}$ OF$=\frac{\sqrt{3}}{2}$,\\
$\therefore$ CD$=2$CF$=\sqrt{3}$, AF$=$OA$+$OF$=\frac{3}{2}$,\\
$\because$ AF$\parallel$AD, and point F is the midpoint of CD,\\
$\therefore$ DE$\perp$CD, AF is the midline of $\triangle CDE$,\\
$\therefore$ DE$=2$AF$=3$,\\
$\therefore$ $S_{\triangle CED}=\frac{1}{2}\times 3\times \sqrt{3}=\boxed{\frac{3\sqrt{3}}{2}}$