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 $4(a^2 + b^2)$ is equal to _______.

Answer 1

$e_E= \sqrt{1− \frac{b ^2}{a^2}}$,$e_H=\sqrt2$ ​

If ⇒ $e_E= \frac{1}{e_H}$

⇒ $\frac{a^2−b^2}{a^2}= \frac{1}{2}$

$2a^{2−2b}2=a^2$

$a^2=2b^2$

and $y= \sqrt\frac{5}{2} x+k$ is tangent to ellipse then

$K^2=a^2× \frac{5}{2}+b^2= \frac{3}{2}$

$6b^2= \frac{3}{2 }⇒ b^2 = \frac{1}{4}$ and $a^2= \frac{1}{2 }$

∴ $4⋅(a^2+b^2)=3$