Proof: Let $M\left(x_{0}, y_{0}\right)$. Then the equation of the line containing the chord of contact $AB$ is
$$\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1.$$

Substituting into the equation of the ellipse $C$ and eliminating $y$ gives
$$\left(1-\frac{x_{0} x}{a^{2}}\right)^{2}=\frac{y_{0}^{2}}{b^{2}}\left(1-\frac{x^{2}}{a^{2}}\right),$$

which simplifies to $\left(\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}\right) x^{2}-2 x_{0} x+a^{2}\left(1-\frac{y_{0}^{2}}{b^{2}}\right)=0$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then
$$k_{M A}=-\frac{b^{2} x_{1}}{a^{2} y_{1}}, k_{M B}=-\frac{b^{2} x_{2}}{a^{2} y_{2}},$$

and $k_{M A} k_{M B}=\frac{b^{4} x_{1} x_{2}}{a^{4} y_{1} y_{2}}=-1$.
Also, $\frac{x_{0} x_{1}}{a^{2}}+\frac{y_{0} y_{1}}{b^{2}}=1, \frac{x_{0} x_{2}}{a^{2}}+\frac{y_{0} y_{2}}{b^{2}}=1$, substituting these gives
$\frac{x_{1} x_{2}}{a^{4}}+\frac{y_{1} y_{2}}{b^{4}}=\frac{x_{1} x_{2}}{a^{4}}+\frac{1}{y_{0}^{2}}\left(1-\frac{x_{0} x_{1}}{a^{2}}\right)\left(1-\frac{x_{0} x_{2}}{a^{2}}\right)=0$,

which simplifies to $\frac{x_{0}^{2}+y_{0}^{2}}{a^{4}} x_{1} x_{2}-\frac{x_{0}}{a^{2}}\left(x_{1}+x_{2}\right)+1=0$.
Using Vieta's formulas, we substitute to get
$$\frac{x_{0}^{2}+y_{0}^{2}}{a^{2}}\left(1-\frac{y_{0}^{2}}{b^{2}}\right)-\frac{2 x_{0}^{2}}{a^{2}}+\left(\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}\right)=0.$$

Simplifying and rearranging gives $x_{0}^{2}+y_{0}^{2}=a^{2}+b^{2}$.
Therefore, the locus of the intersection point $M$ of the two tangents is
$$x^{2}+y^{2}=a^{2}+b^{2}.$$