13.4.33 **The ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has its left vertex at $A(-a, 0)$. Points $M$ and $N$ are on the ellipse $C$, and $M A \perp N A$. Does the line $M N$ pass through a fixed point? If so, find the coordinates of this point; if not, explain the reason.
When $M N \perp x$ axis, we can set $M\left(x_{0}, \frac{b}{a} \sqrt{a^{2}-x_{0}^{2}}\right), N\left(x_{0},-\frac{b}{a} \sqrt{a^{2}-x_{0}^{2}}\right)$. Since $M A \perp N A$, we have $0=\overrightarrow{A M} \cdot \overrightarrow{A N}=\left(x_{0}+a\right)^{2}-\frac{b^{2}}{a^{2}}\left(a^{2}-x_{0}^{2}\right)$, which simplifies to $\left(a^{2}+b^{2}\right) x_{0}^{2}+2 a^{3} x_{0}+a^{2}\left(a^{2}-b^{2}\right)=0$. Therefore, $x_{0}=-\frac{a\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}}$ or $x_{0}=-a$ (discard). Thus, $l_{\mathrm{MN}}: x=\frac{a\left(b^{2}-a^{2}\right)}{a^{2}+b^{2}}$. When $M N$ is not perpendicular to the $x$ axis, we can set $l_{M N}: y=k x+m$, and solve the system of equations
$$
\left\{\begin{array}{l}
y=k x+m, \\
b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}
\end{array} \Rightarrow\left(b^{2}+a^{2} k^{2}\right) x^{2}+2 a^{2} k m x+a^{2}\left(m^{2}-b^{2}\right)=0,\right.
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then
$$
\begin{aligned}
0 & =\overrightarrow{A M} \cdot \overrightarrow{A N}=\left(x_{1}+a\right)\left(x_{2}+a\right)+y_{1} y_{2} \\
& =\left(x_{1}+a\right)\left(x_{2}+a\right)+\left(k x_{1}+m\right)\left(k x_{2}+m\right) \\
& =\left(1+k^{2}\right) x_{1} x_{2}+(a+k m)\left(x_{1}+x_{2}\right)+a^{2}+m^{2} \\
& =\left(1+k^{2}\right) \frac{a^{2}\left(m^{2}-b^{2}\right)}{b^{2}+a^{2} k^{2}}-(a+k m) \frac{2 a^{2} k m}{b^{2}+a^{2} k^{2}}+a^{2}+m^{2},
\end{aligned}
$$

which simplifies to $\left(a^{2}+b^{2}\right) m^{2}-2 a^{3} k m+a^{2} k^{2}\left(a^{2}-b^{2}\right)=0$, solving for $m$ gives $m=a k$ or $m=\frac{a k\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}}$. When $m=a k$, $l_{M N}: y=k x+a k=k(x+a)$, always passing through the point $(-a, 0)$, which does not meet the requirement; when $m=\frac{a k\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}}$, $l_{M N}: y=k x+\frac{a k\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}}=k\left(x+\frac{a\left(a^{2}-b^{2}\right)}{a^{2}+b^{2}}\right)$, always passing through the point $\left(\frac{a\left(b^{2}-a^{2}\right)}{a^{2}+b^{2}}, 0\right)$. In summary, the line $M N$ always passes through the point $\left(\frac{a\left(b^{2}-a^{2}\right)}{a^{2}+b^{2}}, 0\right)$.

Alternatively, we can set $l_{A M}: y=k(x+a), l_{A N}: y=-\frac{1}{k}(x+a)$, then the quadratic curve passing through the intersection points of the ellipse: $b^{2} x^{2}+a^{2} y^{2}-a^{2} b^{2}=0$ and the two lines $A M, A N:[y-k(x+a)]\left[y+\frac{1}{k}(x+a)\right]=0$ is: $b^{2} x^{2}+a^{2} y^{2}-a^{2} b^{2}+\lambda[y-k(x+a)]\left[y+\frac{1}{k}(x+a)\right]=0$.

When this curve is a line passing through $M, N$ and a point $A$, its equation must be in the form: $(x+a) \cdot F(x, y)=0$ (where $F(x, y)=0$ is a linear equation in $x, y$), so $b^{2} x^{2}+\left(a^{2}+\lambda\right) y^{2}-a^{2} b^{2}$ must contain the factor $x+a$, hence $\lambda=-a^{2}$. Thus, the quadratic curve system can be transformed into $(x+a) \left[\left(a^{2}+b^{2}\right) x+a^{2}\left(k-\frac{1}{k}\right) y+a\left(a^{2}-b^{2}\right)\right]=0$. Therefore, the equation of the line $M N$ is: $\left(a^{2}+b^{2}\right) x+a^{2}\left(k-\frac{1}{k}\right) y+a\left(a^{2}-b^{2}\right)=0$, always passing through the point $\left(\frac{a\left(b^{2}-a^{2}\right)}{a^{2}+b^{2}}, 0\right)$.