Question

If $e_1$ and $e_2$ are respectively the eccentricities of the ellipse $\frac{x^2}{18}+\frac{y^2}{4}=1$

and the hyperbola $\frac{x^2}{9}+\frac{y^2}{4}=1$,then

write the value of $2e_1^2+e_2^2.$

Answer 1

$Wehave,$

$\frac{x^2}{9}-\frac{y^2}{4}=1$

$\Rightarrow \frac{x^2}{3^2}-\frac{y^2}{2^2}=1$

$\text{This equation of the hyperbola is of the}$

$form\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1,$

$where,a_2^2=9and{b}_2^2=4$

$\therefore e_2=\sqrt{1+\frac{b_2^2}{a_2^2}}$

$=\sqrt{1+\frac{4}{9}}$

$=\sqrt{13}9$

$=\frac{\sqrt{1}3}{9}$

$Now,$

$2e_1^2+e_2^2=2{\left[\begin{array}{c}\frac{\sqrt{7}}{3}\end{array}\right]}^2+{\left[\begin{array}{c}\frac{\sqrt{1}3}{3}\end{array}\right]}^2$

$=2\times \frac{7}{9}+\frac{13}{9}$

$=\frac{14}{9}+\frac{13}{9}$

$=\frac{14+13}{9}$

$=\frac{27}{9}$

$=3$

$\therefore 2e_1^2+e_2^2=3$