Question 14
Let $a$, $b$, $x$, and $y$ be real numbers with $a \gt 4$ and $b \gt 1$ such that
$$
\frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} = \frac{(x - 20)^2}{b^2 - 1} + \frac{(y - 11)^2}{b^2} = 1.
$$
Find the least possible value for $a + b$.
$$
\frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} = \frac{(x - 20)^2}{b^2 - 1} + \frac{(y - 11)^2}{b^2} = 1.
$$
Find the least possible value for $a + b$.