Example 6. The chord $\mathrm{AB}$ of the ellipse $\frac{x^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ passes through a fixed point $\mathrm{M}\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$, find the locus of the intersection point $\mathrm{P}$ of the two tangents through $\mathrm{A}$ and $\mathrm{B}$.
Let $\mathrm{P}(\overline{\mathrm{x}}, \overline{\mathrm{y}})$, then the equation of $\mathrm{AB}$ is
$$
\frac{\bar{x} x}{a^{2}}+\frac{\bar{y} y}{b^{2}}=1
$$
$\because M \in A B, \quad \therefore \frac{\bar{x} x_{0}}{a^{2}}+\frac{\bar{y} y_{0}}{b^{2}}=1$.
$\therefore$ The locus of point $\mathrm{P}$ is the line: $\frac{x_{0} x}{\mathrm{a}^{2}}+\frac{\mathrm{y}_{0} \mathrm{y}}{\mathrm{b}^{2}}=1$.
In particular, if the fixed point is the focus $\mathrm{F}(\mathrm{c}, 0)$, then the locus of point $\mathrm{P}$ is the directrix $\mathrm{x}=\frac{\mathrm{a}^{2}}{\mathrm{c}}$.