13.7.11 *N* As shown in the figure, from a moving point *P* on the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2} (a > b > 0)$, two tangents *PA* and *PB* are drawn to the circle $x^{2} + y^{2} = b^{2}$. The points of tangency are *A* and *B*. The line *AB* intersects the *x*-axis and *y*-axis at *M* and *N*, respectively.
(1) Find the minimum value of the area of $\triangle M O N$;
(2) Does there exist a point *P* on the ellipse such that the two tangents drawn from *P* to the circle are perpendicular to each other? If so, what conditions must *a* and *b* satisfy, and find the coordinates of point *P*; if not, explain why.
Let the coordinates of point $P$ be $(a \cos \theta, b \sin \theta)$.
(1) The equation of the chord of contact $AB$ determined by point $P$ is $a \cos \theta x + b \sin \theta y = b^2$. Therefore, $|OM| = \frac{b^2}{a \cos \theta}$, $|ON| = \frac{b}{\sin \theta}$, and $S_{\triangle OMN} = \frac{1}{2} |OM| \cdot |ON| = \frac{b^3}{a \sin 2\theta}$. Thus, $\left(S_{\triangle OMN}\right)_{\min} = \frac{b^3}{a}$.
(2) Suppose there exist two tangents such that $PA \perp PB$. Since $OA \perp PA$ and $OB \perp PB$, $OAPB$ forms a square, and $|OP| = \sqrt{2} b$, which means $a^2 \cos^2 \theta + b^2 \sin^2 \theta = 2b^2$, and $\cos^2 \theta = \frac{b^2}{a^2 - b^2}$. Since $\cos^2 \theta \leq 1$, it follows that $a^2 \geq 2b^2$, or $a \geq \sqrt{2} b$. When $a = \sqrt{2} b$, point $P$ is $(a, 0)$ or $(-a, 0)$; when $a > \sqrt{2} b$, point $P$ is
$$
\left(\frac{ab}{\sqrt{a^2 - b^2}}, \pm \frac{b \sqrt{a^2 - 2b^2}}{\sqrt{a^2 - b^2}}\right) \text{ or } \left(-\frac{ab}{\sqrt{a^2 - b^2}}, \pm \frac{b \sqrt{a^2 - 2b^2}}{\sqrt{a^2 - b^2}}\right);
$$
When $0 < b < a < \sqrt{2} b$, there is no point $P$ on the ellipse that meets the requirements of the problem.