11. (20 points) How many inscribed isosceles right triangles $\triangle ABC$ with the right angle vertex $A(0,1)$ are there on the ellipse $\frac{x^{2}}{a^{2}}+y^{2}=1$?
11. Let $l_{A B}: y=k x+1(k>0)$.

Substituting into the ellipse equation, we get
$$
\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0 \text {. }
$$

Thus, $x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$,
$$
|A B|=\sqrt{1+k^{2}}\left|x_{A}-x_{B}\right|
$$
$$
=\frac{2 a^{2} k \sqrt{1+k^{2}}}{1+a^{2} k^{2}} \text {. }
$$

Similarly, $|A C|=\frac{2 a^{2} \sqrt{1+k^{2}}}{k^{2}+a^{2}}$.
From $|A B|=|A C|$, we get
$$
k^{3}-a^{2} k^{2}+a^{2} k-1=0,
$$

which is $(k-1)\left[k^{2}+\left(1-a^{2}\right) k+1\right]=0$.
For the second factor,
$$
\Delta=\left(1-a^{2}\right)^{2}-4=\left(a^{2}-3\right)\left(a^{2}+1\right) \text {. }
$$

Therefore, when $a^{2}>3$, the equation has two positive roots, neither of which is 1; when $a^{2}=3$, the equation has a double root of 1; when $a^{2}<3$, the equation has no real roots. Hence, when $a^{2}>3$, there are 3 triangles that satisfy the condition; when $a^{2} \leqslant 3$ and $a^{2} \neq 1$, there is 1 triangle that satisfies the condition.