Example 10. Derive the equation of a parabola symmetric with respect to the $O x$ axis, with its vertex at the origin, given that the parabola passes through the point $A(2,2)$.

Find the length of the chord passing through the point $M(8,0)$ and inclined to the $O x$ axis at an angle of $60^{\circ}$.
Solution. 1) Let's make a drawing (Fig. 2.27).

2) The canonical equation of such a parabola is $y^{2}=2 p x$. The unknown parameter $p$ will be determined from the condition that the parabola passes through the point $A(2,2)$:

$$
2^{2}=2 \cdot p \cdot 2 \Rightarrow p=1
$$

Thus, the equation of the parabola is $y^{2}=2 x$.

3) Let's find the coordinates of points $B_{1}\left(x_{1}, y_{1}\right)$ and $B_{2}\left(x_{2}, y_{2}\right)$; points $B_{1}$ and $B_{2}$ lie on the parabola, so $y_{1}^{2}=2 x_{1}$ and $y_{2}^{2}=2 x_{2}$. From the right triangles $M B_{2} C_{2}$ and $M B_{\mathrm{I}} C_{\mathrm{I}}$, we have respectively: $y_{2}=\left(x_{2}-8\right) \operatorname{tg} 60^{\circ}=\sqrt{3}\left(x_{2}-8\right)$ and $y_{1}=-\sqrt{3}\left(8-x_{1}\right)$. Thus, the unknown coordinates of points $B_{\mathrm{I}}$ and $B_{2}$ satisfy the systems

$$
\left\{\begin{array} { l } 
{ y _ { 1 } ^ { 2 } = 2 x _ { 1 } } \\
{ y _ { 1 } = - \sqrt { 3 } ( 8 - x _ { 1 } ) }
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
y_{2}^{2}=2 x_{2} \\
y_{2}=\sqrt{3}\left(x_{2}-8\right)
\end{array}\right.\right.
$$

Solving these, we find $B_{1}(6,-2 \sqrt{3})$ and $B_{2}\left(\frac{32}{3} ; \frac{8}{\sqrt{3}}\right)$. The length of the chord $B_{1} B_{2}=\sqrt{\left(\frac{32}{3}-6\right)^{2}+\left(\frac{8}{\sqrt{3}}+2 \sqrt{3}\right)^{2}}=\frac{28}{3}$.

$$
\text { Answer. } B_{1} B_{2}=\frac{28}{3} \text {. }
$$