Five. (Full marks 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the positive direction of the $y$-axis at point $B$. Find the number of isosceles right triangles inscribed in the ellipse with point $B$ as the right-angle vertex.

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Please note that the translation preserves the original format and line breaks.
Let the right-angled isosceles triangle inscribed in the ellipse be $\triangle A B C$, and suppose the equation of $A B$ is
\[
\left\{\begin{array}{l}
x=t \cos \alpha, \\
y=b+t \sin \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (1) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+2 a^{2} b t \sin \alpha=0.
\]
$t=0$ corresponds to point $B$, which we discard. Therefore,
\[
|A B|=-t=\frac{2 a^{2} b \sin \alpha}{b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}.
\]
Suppose the equation of $B C$ is
\[
\left\{\begin{array}{l}
x=t \cos \left(\alpha+90^{\circ}\right)=-t \sin \alpha, \\
y=b+t \sin \left(\alpha+90^{\circ}\right)=b+t \cos \alpha .
\end{array}\right.
\]
(t is a parameter)
Substituting (3) into $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, we get
\[
\left(b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha\right) t^{2}+2 a^{2} b t \cos \alpha=0.
\]
$t=0$ corresponds to point $B$, which we discard. Therefore,
\[
|B C|=-t=\frac{2 a^{2} b \cos \alpha}{b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha}.
\]
Since $|B A|=|B C|$, from (2) and (4) we have
\[
\frac{2 a^{2} b \sin \alpha}{b^{2} \cdot \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha}=\frac{2 a^{2} b \cos \alpha}{b^{2} \sin ^{2} \alpha+a^{2} \cos ^{2} \alpha}.
\]

Rearranging the above equation, we get
\[
(\tan \alpha-1)\left[b^{2} \tan ^{2} \alpha+\left(b^{2}-a^{2}\right) \tan \alpha+b^{2}\right]=0.
\]
From $\tan \alpha-1=0$, we have $\alpha=45^{\circ}$, indicating that $\triangle A B C$ is an isosceles right-angled triangle symmetric about the y-axis;

From $b^{2} \tan ^{2} \alpha+\left(b^{2}-a^{2}\right) \tan \alpha+b^{2}=0$, the discriminant $\Delta=\left(b^{2}-a^{2}\right)^{2}-4 b^{4}=\left(a^{2}-3 b^{2}\right)\left(a^{2}+b^{2}\right)$. When $\Delta>0$, i.e., $a>\sqrt{3} b$, there are two such triangles; when $\Delta=0$, i.e., $a^{2}=3 b^{2}$, we have $\tan \alpha=1$.

Therefore, when $a>\sqrt{3} b$, there are two such triangles; when $b<a \leqslant \sqrt{3} b$, there is one such triangle.