Question 12
Let $a,$ $b,$ $x,$ and $y$ be real numbers with $a \gt 4$ and $b \gt 1$ such that $$
\begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} &= \frac{(x - 20)^2}{b^2 - 1} \\ &\quad {}+ \frac{(y - 11)^2}{b^2} \\ &= 1. \end{aligned}
$$ Find the least possible value of $a + b.$
\begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} &= \frac{(x - 20)^2}{b^2 - 1} \\ &\quad {}+ \frac{(y - 11)^2}{b^2} \\ &= 1. \end{aligned}
$$ Find the least possible value of $a + b.$