13. Let the intersection locus of two perpendicular tangents to the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$ be $C$. Suppose the tangents $P A, P B$ to curve $C$ intersect at point $P$, and are tangent to $C$ at points $A, B$ respectively. Find the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$.
Let $C(m, n)$, the equation of the tangent line to the ellipse is $y-n=k(x-m) \Rightarrow y=k x-(k m-n)$,
Substituting into the ellipse equation yields $\left(5 k^{2}+4\right) x^{2}-10 k(k m-n) x+5(k m-n)^{2}-20=0$.
Let $\Delta=100 k^{2}(k m-n)^{2}-4\left(5 k^{2}+4\right)\left[5(k m-n)^{2}-20\right]=0$
$\Rightarrow\left(m^{2}-5\right) k^{2}-2 k m n+n^{2}-4=0$, since the two tangents to the ellipse are perpendicular,
then $k_{1} k_{2}=\frac{n^{2}-4}{m^{2}-5}=-1 \Rightarrow m^{2}+n^{2}=9$, which means the curve $C: x^{2}+y^{2}=9$.
Let $|P O|=x, \angle A P O=\theta$, then $\cos ^{2} \theta=\frac{|P A|^{2}}{|P O|^{2}}=\frac{x^{2}-9}{x^{2}}$
$\Rightarrow \cos \angle A P B=2 \cos ^{2} \theta-1=\frac{x^{2}-18}{x^{2}}$.
Thus, $\overrightarrow{P A} \cdot \overrightarrow{P B}=|P A|^{2} \cos \angle A P B=\frac{\left(x^{2}-9\right)\left(x^{2}-18\right)}{x^{2}}=x^{2}+\frac{18 \cdot 9}{x^{2}}-27$ $\geqslant 2 \sqrt{18 \cdot 9}-27=18 \sqrt{2}-27$, equality holds when $x^{2}=9 \sqrt{2}$.
In conclusion, the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $18 \sqrt{2}-27$.