JEE PYQ: Ellipse Question 6
Question 6 - 2020 (03 Sep Shift 2)
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) $(8, 12)$ (b) $\left(\frac{20}{3}, 12\right)$ (c) $\left(\frac{24}{5}, 10\right)$ (d) $(8, 10)$
Show Answer
Answer: (d) $(8, 10)$
Solution
$e_1 = \sqrt{1 - \frac{b^2}{25}}$, $e_2 = \sqrt{1 + \frac{b^2}{16}}$
$(e_1 e_2)^2 = 1 \Rightarrow \left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1$
$\Rightarrow b^2 = 9$
$e_1 = \frac{4}{5}$, $e_2 = \frac{5}{4}$
$\alpha = 2(5)(\frac{4}{5}) = 8$, $\beta = 2(4)(\frac{5}{4}) = 10$
$\therefore (\alpha, \beta) = (8, 10)$
BETA
