For the hyperbola $H: x^{2}-y^{2}=1$ and the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,

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 (i) eccentricity of $E$ be reciprocal of the eccentricity of $H$, and (ii) 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

Professor PC Thomas Classes · Answer

(3) : Given\ $H: x^{2}-y^{2}=1, E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b>0$\ $e_{H}=\sqrt{2}, e_{E}=\sqrt{1-\frac{b^{2}}{a^{2}}}$\ We have, $e_{E}=\frac{1}{e_{H}}$\ $\Rightarrow \frac{a^{2}}{a^{2}-b^{2}}=2 \Rightarrow a^{2}=2 b^{2}$\ Since, $y=\sqrt{\frac{5}{2}} x+K$ is common tangent to ellipse and hyperbola, then\ $K^{2}=a^{2} m^{2}+b^{2}=a^{2}\left(\frac{5}{2}ight)+b^{2}=\frac{3}{2}$\ Put $a^{2}=2 b^{2}$, we get $b^{2}=\frac{1}{4} \quad 	herefore \quad a^{2}=\frac{1}{2}$\ Now, $4\left(a^{2}+b^{2}ight)=3$.\