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 $2 ae$
$\therefore $ distance for ellipse $=2 \times 5 \times \frac{4}{5}=8=\alpha$
distance for hyperbola $=2 \times 4 \times \frac{5}{4}=10=\beta$
$\therefore (\alpha, \beta) \equiv(8,10)$