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 \(a^2= \frac{1}{2 }\)

∴ \(4⋅(a^2+b^2)=3\)