20. Let the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1), \operatorname{Rt} \triangle A B C$ with $A(0,1)$ as the right-angle vertex, and sides $A B, B C$ intersecting the ellipse at points $B, C$. If the maximum area of $\triangle A B C$ is $\frac{27}{8}$, find the value of $a$.
20. Let $l_{A B}: y=k x+1(k>0)$. Then $l_{A C}: y=-\frac{1}{k} x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ \frac{x^{2}}{a^{2}}+y^{2}=1,\end{array}\right.$,
$\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0$
$\Rightarrow x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$.
Thus, $|A B|=\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2} k^{2}}$.
Similarly, $|A C|=\sqrt{1+\frac{1}{k^{2}}} \cdot \frac{2 a^{2} k}{a^{2}+k^{2}}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}|A B||A C|$
$=2 a^{4} \frac{k\left(1+k^{2}\right)}{\left(1+a^{2} k^{2}\right)\left(a^{2}+k^{2}\right)}$
$=2 a^{4} \frac{k+\frac{1}{k}}{a^{2}\left(k^{2}+\frac{1}{k^{2}}\right)+a^{4}+1}$.
Let $t=k+\frac{1}{k} \geqslant 2$. Then
$S_{\triangle A B C}=\frac{2 a^{4} t}{a^{2} t^{2}+\left(a^{2}-1\right)^{2}}$
$=\frac{2 a^{4}}{a^{2} t+\frac{\left(a^{2}-1\right)^{2}}{t}}$.
And $a^{2} t+\frac{\left(a^{2}-1\right)^{2}}{t} \geqslant 2 a\left(a^{2}-1\right)$, when $t=\frac{a^{2}-1}{a}$, equality holds.
Thus, when $t=\frac{a^{2}-1}{a}$, $\left(S_{\triangle A B C}\right)_{\max }=\frac{a^{3}}{a^{2}-1}$.
From $\frac{a^{3}}{a^{2}-1}=\frac{27}{8}$
$\Rightarrow(a-3)\left(8 a^{2}-3 a-9\right)=0$
$\Rightarrow a=3, a=\frac{3+\sqrt{297}}{16}$.
From $\frac{a^{2}-1}{a}>2 \Rightarrow a>1+\sqrt{2}$.
Therefore, $a=\frac{3+\sqrt{297}}{16}$ is not valid, and is discarded.
Thus, $a=3$.