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 distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha ,\beta )$ is equal to :

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

Answer 1

Answer: (A)

For ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1(b<5)$
Let $e_1$ is eccentricity of ellipse
$\therefore b^2=25\left(1-e_1^2\right)$..... (1)
Again for hyperbola
$\frac{x^2}{16}-\frac{y^2}{b^2}=1$
Let $e_2$ is eccentricity of hyperbola
$\therefore b^2=16\left(e_2^2-1\right)$.....(2)
by (1) & (2)
$25\left(1-e_1^2\right)=16\left(e_2^2-1\right)$
Now $e_1\cdot e_2=1$ (given)
$\therefore 25\left(1-e_1^2\right)=16\left(\frac{1-e_1^2}{e_1^2}\right)$
or $e_1=\frac{4}{5}$
$\therefore e_2=\frac{5}{4}$
Now distance between foci is $2ae$
$\therefore$ distance for ellipse $=2\times 5\times \frac{4}{5}=8=\alpha$
distance for hyperbola $=2\times 4\times \frac{5}{4}=10=\beta$
$\therefore (\alpha ,\beta )\equiv (8,10)$