Example 9. Derive the equation of a parabola symmetric with respect to the $O y$ axis, given that it passes through the points of intersection of the line $x-y=0$ and the circle $x^{2}+y^{2}-6 y=0$.
Solution. The equation of the desired parabola must have the form $x^{2} = 2 p y$; it is depicted in Fig. 2.26. Let's find the points of intersection of the given line and circle:

\[
\begin{aligned}
& \left\{\begin{array}{l}
y=x \\
x^{2}+y^{2}-6 y=0
\end{array}\right. \\
& 2 x^{2}-6 x=0 ; x_{1}=0, x_{2}=3
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-030.jpg?height=474&width=473&top_left_y=181&top_left_x=1409)

Fig. 2.26

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-030.jpg?height=343&width=477&top_left_y=312&top_left_x=1956)

Fig. 2.27

We obtained $M_{1}(0,0)$ and $M_{2}(3,3)$. Since the point $M_{2}(3,3)$ lies on the parabola, the equality $3^{2}=2 p \cdot 3 \Rightarrow p=\frac{3}{2}$ holds, and the desired equation of the parabola is $x^{2}=3 y$.