Question

Let $e_1$ and $e_2$ be the eccentricities of the ellipse, $\frac{x^2}{25}+\frac{y^2}{b^2}=1(b<5)$ and the hyperbola, $\frac{x^2}{16}-\frac{y^2}{b^2}=1$ respectively satisfying $e_1{e}_2=1$. If $\alpha$ and $\beta$ are the distance between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha ,\beta )$ is equal to:

• A. $(8,12)$
• B. $(\frac{24}{5},10)$
• C. $(\frac{20}{3},12)$
• D. $(8,10)$

Answer 1

The correct option is D $(8,10)$

$e_1=\sqrt{1-\frac{b^2}{25}},e_2=\sqrt{1+\frac{b^2}{16}}$
$e_1{e}_2=1$
$\Rightarrow {\left(e_1{e}_2\right)}^2=1$
$\Rightarrow (1-\frac{b^2}{25})(1+\frac{b^2}{16})=1$
$\Rightarrow 1+\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=1$
$\Rightarrow \frac{9b^2}{16\cdot 25}-\frac{b^4}{400}=0$
$\Rightarrow 9b^2-b^4=0$
$\Rightarrow b^2(9-b^2)=0,b\ne 0$
$\Rightarrow 9-b^2=0$
$\therefore b^2=9$
$\Rightarrow e_1=\frac{4}{5}$
$\Rightarrow e_2=\frac{5}{4}$
$\Rightarrow \alpha =2a{e}_1=2\left(5\right)\left(e_1\right)=8$
$\Rightarrow \beta =2a{e}_2=2\left(4\right)\left(e_2\right)=10$
$\therefore (\alpha ,\beta )=(8,10)$