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

Answer: 3

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