Question

For the hyperbola $H : x ^2- y ^2=1$ and the ellipse $E : \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1, a > b >0$, let the
(1) eccentricity of $E$ be reciprocal of the eccentricity of $H$, and
(2) the line $y=\sqrt{\frac{5}{2}} x+K$ be a common tangent of $E$ and $H$.
Then $4\left(a^2+b^2\right)$ is equal to

Answer 1

$e _{ E }=\sqrt{1-\frac{ b ^2}{ a ^2}}, e _{ H }=\sqrt{2} $
$\text { If } \Rightarrow e _{ E }=\frac{1}{ e _{ H }} $
$ \Rightarrow \frac{ a ^2- b ^2}{ a ^2}=\frac{1}{2}$
$ 2 a ^{2-2 b} 2= a ^2 $
$ a ^2=2 b ^2$
and $y =\sqrt{\frac{5}{2}} x + k$ is tangent to ellipse then
$K ^2= a ^2 \times \frac{5}{2}+ b ^2=\frac{3}{2}$
$6 b^2=\frac{3}{2} \Rightarrow b^2=\frac{1}{4}$ and $a^2=\frac{1}{2}$
$\therefore 4 \cdot\left(a^2+b^2\right)=3$