Question

Let $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, a $>0,b>0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2\sqrt{2}+\sqrt{14})$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then value of $a^2+b^2$ is equal to _______

Answer 1

Answer: 88

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given $e^2=1+\frac{b^2}{a^2}\Rightarrow \frac{11}{4}=1+\frac{b^2}{a^2}\Rightarrow b^2=\frac{7}{4}{a}^2$
$\therefore \frac{x^2}{(a{)}^2}-\frac{y^2}{{\left(\frac{\sqrt{7}}{2}a\right)}^2}=1$ Now given
$2a+2\cdot \frac{\sqrt{7}a}{2}=4(2\sqrt{2}+\sqrt{14})$
$a(2+\sqrt{7})=4\sqrt{2}(2+\sqrt{7})$
$a=4\sqrt{2}\Rightarrow a^2=32$
$b^2=\frac{7}{4}\times 16\times 2=56$