Let a, b, x and y be real numbers with a > 4 and b > 1 such that $\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{{(x-20)}^2}{b^2-1}+\frac{{(y-11)}^2}{b^2}=1$ the least possible value of a + b is_______

$\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=1$

Centre (0, 0) foci (-4, 0) and (4, 0)

$\frac{{(x-20)}^2}{b^2-1}+\frac{{(y-11)}^2}{b^2}=1$

Centre: (20, 11) foci (20, 10) and (20, 12)

$2a=S_{1}P+S_{2}P$

$2b=S_{3}P+S_{4}P$

$2a+2b=S_{1}P+S_{2}P+S_{3}P+S_{4}P$

To make the sum minimum, P is point of intersection of the line that passes through (-4,0) and (20,10), and the line that passes through (4, 0) and (20, 12)

$\Rightarrow S_{1}P+S_{3}P=S_2{S}_3$

$S_{1}P+S_{4}P=S_1{S}_4$

$2a+2b=26+20\Rightarrow a+b=23$