10. $P$ is any point on the ellipse $\frac{(x+4)^{2}}{9}+\frac{y^{2}}{16}=1$, and $PA, PB$ are two tangent lines to the parabola $y^{2}=4x$, with $A, B$ being the points of tangency. Find the maximum area of $\triangle PAB$, and determine the coordinates of point $P$ when the area of $\triangle PAB$ is maximized.
10. Solution: Let the coordinates of point $P$ be $\left(x_{0}, y_{0}\right)$, and the coordinates of the tangent points $A$ and $B$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, respectively. Then the equation of the tangent line $A P$ is $y_{1} y=2\left(x+x_{1}\right)$, and the equation of the tangent line $B P$ is $y_{2} y=2\left(x+x_{2}\right)$.
$\because$ Point $P\left(x_{0}, y_{0}\right)$ lies on the tangent lines $A P$ and $B P$, $\therefore y_{1} y_{0}=2\left(x_{0}+x_{1}\right), y_{2} y_{0}=2\left(x_{0}+x_{2}\right)$. These two equations indicate that the line $y_{0} y=2\left(x+x_{0}\right)$ passes through points $A$ and $B$, i.e., the equation of line $A B$ is $y_{0} y=2\left(x+x_{0}\right)$.
Let the line through $P$ parallel to the $x$-axis intersect line $A B$ at $D$. Substituting the $y$-coordinate of $B$, $y_{0}$, into $y_{0} y=2\left(x+x_{0}\right)$, we get $x=\frac{y_{0}^{2}}{2}-x_{0}$. Therefore, $|P D|=\left(\frac{y_{0}^{2}}{2}-x_{0}\right)-x_{0}=\frac{y_{0}^{2}-4 x_{0}}{2}$.
From the system of equations $\left\{\begin{array}{l}y^{2}=4 x \\ y_{0} y=2\left(x+x_{0}\right)\end{array}\right.$, eliminating $x$ yields $y^{2}-2 y_{0} y+4 x_{0}=0$,
By Vieta's formulas, we have $y_{1}+y_{2}=2 y_{0}, y_{1} y_{2}=4 x_{0}$,
Thus, $\left|y_{1}-y_{2}\right|=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}=\sqrt{4 y_{0}^{2}-16 x_{0}}=2 \sqrt{y_{0}^{2}-4 x_{0}}$.
$\therefore S_{\triangle P A B}=\frac{1}{2}|P D|\left|y_{1}-y_{2}\right|=\frac{1}{2}\left(y_{0}^{2}-4 x_{0}\right)^{\frac{3}{2}}$.
Let $P(-4+3 \cos \theta, 4 \sin \theta)(0 \leqslant \theta<2 \pi)$, substituting into the above expression, we get
$$
S_{\triangle P A B}=\frac{1}{2}\left(16 \sin ^{2} \theta-12 \cos \theta+16\right)^{\frac{3}{2}}=4\left[-4\left(\cos \theta+\frac{3}{8}\right)^{2}+\frac{137}{16}\right]^{\frac{3}{2}} \text {. }
$$

Since $y=u^{\frac{3}{2}}$ is an increasing function (where $u=-4\left(\cos \theta+\frac{3}{8}\right)^{2}+\frac{137}{16}$), when $\cos \theta=-\frac{3}{8}$, $u$ reaches its maximum value $\frac{137}{16}$, at which point $S_{\triangle P A B}$ reaches its maximum value $\frac{137 \sqrt{137}}{16}$, and the coordinates of point $P$ are $\left(-\frac{41}{8}, \frac{\sqrt{55}}{2}\right)$ or $\left(-\frac{41}{8},-\frac{\sqrt{55}}{2}\right)$.