10. Taking the vertex $C(0,1)$ of the ellipse $x^{2}+a^{2} y^{2}=a^{2}(a>1)$ as the right-angle vertex, construct an inscribed isosceles right-angled triangle $A B C$ in this ellipse. The maximum number of such triangles that can be constructed is $\qquad$.
10. 3

Obviously, such a triangle must exist. Without loss of generality, assume points $A$ and $B$ are located on the left and right sides of the $y$-axis, respectively. Let the slope of $CA$ be $k (k > 0)$, and the equation of the line $CA$ be $y = kx + 1$. Substituting this into the ellipse equation yields $\left(a^{2} k^{2} + 1\right) x^{2} + 2 a^{2} k x = 0$. This gives $x = 0$ or $x = \frac{2 a^{2} k}{a^{2} k^{2} + 1}$. Therefore, the coordinates of $A$ are
$$
\left(-\frac{2 a^{2} k}{a^{2} k^{2} + 1}, 0\right), \text{ hence } |CA| = \frac{2 a^{2} k \sqrt{1 + k^{2}}}{a^{2} k^{2} + 1}.
$$

Similarly, we get $|CB| = \frac{2 a^{2} \sqrt{1 + k^{2}}}{a^{2} + k^{2}}$.
Using $|CA| = |CB|$, we obtain $(k - 1)\left[k^{2} - \left(a^{2} - 1\right) k + 1\right] = 0$, which gives $k = 1$ or $k^{2} - \left(a^{2} - 1\right) k + 1 = 0$. For the latter, when $1 < a < \sqrt{3}$, there are two distinct real roots.