Example 3. If the ellipse $(x-a)^{2}+2 y^{2}=1$ intersects the parabola $y$ $=2 \mathrm{px}$, what relationship must $\mathrm{z}$ and $\mathrm{p}$ satisfy so that the tangents to the ellipse and the parabola at the intersection points are perpendicular to each other?
Solution: Obviously, the figure is symmetric about the x-axis, and the two intersection points are respectively in the
I and IV quadrants. Let the intersection point in the first quadrant be $\left(x_{0}, y_{0}\right)$. Then, from the ellipse equation $(x-a)^{2}+2 y^{2}=1$, we get $y'=\frac{-(x-a)}{2 y}$. Therefore, at the point $\left(x_{0}, y_{0}\right)$, the slope of the tangent line is
$k_{1}=\frac{-(x_{0}-a)}{2 y_{0}}$. Also, from the parabola equation $y^{2}=2 p x$, we get $y'_{x}=\frac{p}{y}$. Therefore, at the point $\left(x_{0}, y_{0}\right)$, the slope of the tangent line is $k_{2}=\frac{p}{y_{0}}$. By the condition that the tangents are perpendicular, we get
$$
\frac{-(x_{0}-a)}{2 y_{0}} \cdot \frac{p}{y_{0}}=-1,
$$

Also, from the intersection of the two curves at $\left(x_{0}, y_{0}\right)$, we get
$$
\begin{array}{l}
\left(x_{0}-a\right)^{2}+2 y_{0}^{2}=1, \\
y_{0}^{2}=2 p x_{0} .
\end{array}
$$

From equations (8) and (10), we easily get
$$
a=-3 x_{0} \text{. }
$$

Substituting equations (10) and (11) into equation (9), we get
$$
\begin{array}{c}
16 x_{0}^{2}+4 p x_{0}-1=0 \\
\therefore \quad x_{0}=\frac{\pm p^{2}+4}{3},
\end{array}
$$

But the point $\left(x_{0}, y_{0}\right)$ is in the first quadrant, so $x_{0}>0$,
$$
\therefore \quad x_{0}=\frac{\sqrt{p^{2}+4}-p}{8} .
$$

From equation (11), we have
$$
a=\frac{-3}{8}\left(\sqrt{p^{2}+4}-p\right),
$$

which gives $16 a^{2}-12 a p-9=0, (a \neq 0)$, this is the relationship that $a$ and $p$ should satisfy.

This example is a more general case of the 1980 college entrance examination question 9. The 1980 exam question 9 gave the parabola $y^{2}=2 x$ and the circle $(x-a)^{2}+y^{2}=1$, and used the condition that the tangents at the intersection point are perpendicular to find the value of $a$. Since the figures involved in the problem are a circle and a parabola, it can be solved using geometric methods or derivatives. However, if we use geometric methods to solve the more general problem we have developed, it would be quite difficult. This further illustrates the convenience of using derivatives.