Example 2. A moving point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ draws tangents to the ellipse $\mathrm{b}^{2} \mathrm{x}^{2}$ with angles $\theta_{1}, \theta_{2}$. (1) When $\operatorname{tg} \theta_{1}+\operatorname{tg} \theta_{2}=\mathrm{m}$ (a constant), find the equation of the locus of point $P$; (2) When $\operatorname{ctg} \theta_{1} +\operatorname{ctg} \theta_{2}=\mathrm{n}$ (a constant), find the equation of the locus of point $\mathrm{P}$.
(1) Let the slope of a tangent be $k_{1}$, then the tangent line is
$$
\begin{aligned}
y & =k_{1} x+\sqrt{k_{1}^{2} a^{2}+b^{2}} \\
\text { or } y & =k_{1} x-\sqrt{k_{1}^{2} a^{2}+b^{2}},
\end{aligned}
$$

i.e., $\pm \sqrt{k_{1}^{2} a^{2}+b^{2}}=y-k_{1} x$. Squaring both sides, we get:
$$
a^{2} k_{1}^{2}+b^{2}=y^{2}-2 x y k_{1}+k_{1}^{2} x^{2},
$$

i.e., $\left(a^{2}-x^{2}\right) k_{1}^{2}+2 x y k_{1}+\left(b^{2}-y^{2}\right)=0$.
Similarly, if the slope of another tangent is $k_{2}$, then we must have:
$$
\left(a^{2}-x^{2}\right) k_{2}^{2}+2 x y k_{2}+\left(b^{2}-y^{2}\right)=0.
$$

From (A) and (B), we know that $k_{1}, k_{2}$ are the two roots of the equation $\left(a^{2}-x^{2}\right) k^{2}+2 x y k+\left(b^{2}-y^{2}\right)=0$,
thus $k_{1}+k_{2}=\frac{-2 x y}{a^{2}-x^{2}}$.
Given that $\theta_{1}+\operatorname{tg} \theta_{2}=m$,
$$
\therefore \frac{-2 x y}{a^{2}-x^{2}}=m,
$$

i.e., $m\left(x^{2}-a^{2}\right)=2 x y$ is the required equation.
(2) From $\left(a^{2}-x^{2}\right) k^{2}+2 x y k+\left(b^{2}-y^{2}\right)=0$,

we get $k_{1}+k_{2}=\frac{-2 x y}{a^{2}-x^{2}}, k_{1} k_{2}=\frac{b^{2}-y^{2}}{a^{2}-x^{2}}$.
Given the condition $\operatorname{ctg} \theta_{1}+\operatorname{ctg} \theta_{2}=n$,
$$
\begin{array}{l}
\text { i.e., } \frac{1}{k_{1}}+\frac{1}{k_{2}}=n, \frac{k_{1}+k_{2}}{k_{1} k_{2}}=n, \\
\therefore \frac{-2 x y}{b^{2}-y^{2}}=n, \text { i.e., } n\left(y^{2}-b^{2}\right)=2 x y
\end{array}
$$

is the required equation.