Mathematics - Hyperbola Question with Solution | TestHub

Given

$E\equiv 3x^2+4y^2=12 \Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1$

Now,  $e_E=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$

Length of L.R. $=\frac{2b^2}{a}=\frac{2\times 3}{2}=3$

Now $H\equiv \frac{x^2}{a^2}-\frac{y^2}{1}=1$

 Length of L.R.$=\frac{2b^2}{a}=\frac{2\times 1}{a}$

Given length of L.R. of $E$ and $H$ are equal

So $\frac{2}{a}=3 \Rightarrow a=\frac{2}{3}$

Now $e_H=\sqrt{\frac{b^2}{a^2}+1}=\sqrt{\frac{1}{{\left({\displaystyle \frac{2}{3}}\right)}^2}}+1 \Rightarrow e_H=\sqrt{\frac{9}{4}+1}=\sqrt{\frac{13}{4}}$

So $12\left(e_H^2+e_E^2\right)=12\left(\frac{13}{4}+\frac{1}{4}\right)=12\left(\frac{14}{4}\right)=14\times 3=42$