Example 2. Given, a moving point $\mathrm{P}(\mathrm{X}, \mathrm{Y})$ draws two tangent lines to the parabola $\dot{y}^{2}=2 \mathrm{ps}(\mathrm{p}>0)$ which are perpendicular to each other, find the equation of the locus of point $\mathrm{P}$.

 

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$$
\left(x_{1}, y_{1}\right) 、\left(x_{2}, y_{2}\right),
$$

It is known that neither of the two tangent points is the origin (Figure 1).
From $y^{2}=2 p x$, we get $2 y y^{\prime} = 2 p, y^{\prime}=\frac{p}{y}$. Since the tangent line passes through point $P(x, y)$, from equation (4), we have
$$
\left\{\begin{array}{l}
\frac{Y - y_{1}}{X - x_{1}} = \frac{p}{y_{1}}, \\
y_{1}^{2} = 2 p x_{1};
\end{array} \left\{\begin{array}{l}
\frac{Y - y_{2}}{X - x_{2}} = \frac{p}{y_{2}}, \\
y_{2}^{2} = 2 p x_{2}.
\end{array}\right.\right.
$$

By eliminating $x_{1}$ and $x_{2}$ from the above two sets of equations, we get
$$
\begin{array}{l}
y_{1}^{2} - 2 Y y_{1} + 2 p X = 0, \\
y_{2}^{2} - 2 Y y_{2} + 2 p X = 0.
\end{array}
$$

Thus, $y_{1}$ and $y_{2}$ are the two roots of the quadratic equation $y^{2} - 2 Y y + 2 p X = 0$. Solving this equation, we get
$$
\begin{aligned}
y_{1,2} & = \frac{2 Y \pm \sqrt{4 Y^{2} - 8 p X}}{2} \\
& = Y \pm \sqrt{Y^{2} - 2 p X}.
\end{aligned}
$$

From the above, we know that when and only when $Y^{2} > 2 p X$, the equation has two real roots, and point $P(X, Y)$ can draw two tangents to $y^{2} = 2 p x$. Since the two tangents are perpendicular to each other, we have
$$
k_{1} \cdot k_{2} = \frac{p}{y_{1}} \cdot \frac{p}{y_{2}} = -1,
$$

which gives $y_{1} \cdot y_{2} = -p^{2}$,
i.e., $\left(Y + \sqrt{Y^{2} - 2 p X}\right)\left(Y - \sqrt{Y^{2} - 2 p X}\right)$
$$
= 2 p X = -p^{2}.
$$
$\therefore X = -\frac{p}{2}$ is the equation of the locus of point $P$.