A hyperbola touches y – axis and has its centre and one of the foci are at $(\frac{5}{2},20)$  and (10, 24) respectively; then length of its transverse axis is

We have ae =  $\sqrt{{(10-\frac{5}{2})}^2+{(24-20)}^2}=\sqrt{\frac{225}{4}+6}=\frac{17}{2}$
Clearly other focus is at (-5, 16)
As hyperbola touches y – axis $\Rightarrow$  y – axis is tangent
$\Rightarrow p_1{p}_2=b^2\Rightarrow b^2=50=a^2{e}^2-a^2\Rightarrow 50=\frac{289}{4}-a^2\Rightarrow a^2-\frac{89}{4}\Rightarrow 2a=\sqrt{89}$