JEE PYQ: Ellipse Question 5
Question 5 - 2020 (02 Sep Shift 2)
For some $\theta \in \left(0, \frac{\pi}{2}\right)$, if the eccentricity of the hyperbola, $x^2 - y^2 \sec^2 \theta = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^2 \sec^2 \theta + y^2 = 5$, then the length of the latus rectum of the ellipse, is:
(a) $2\sqrt{6}$ (b) $\sqrt{30}$ (c) $\frac{2\sqrt{5}}{3}$ (d) $\frac{4\sqrt{5}}{3}$
Show Answer
Answer: (d) $\frac{4\sqrt{5}}{3}$
Solution
Hyperbola: $\frac{x^2}{10} - \frac{y^2}{10\cos^2\theta} = 1 \Rightarrow e_1 = \sqrt{1 + \cos^2\theta}$
Ellipse: $\frac{x^2}{5\cos^2\theta} + \frac{y^2}{5} = 1$
$\Rightarrow e_2 = \sqrt{1 - \cos^2\theta} = \sin\theta$
According to the question, $e_1 = \sqrt{5}e_2$
$\Rightarrow 1 + \cos^2\theta = 5\sin^2\theta \Rightarrow \cos^2\theta = \frac{2}{3}$
Now length of latus rectum of ellipse $= \frac{2a^2}{b} = \frac{10\cos^2\theta}{\sqrt{5}} = \frac{20}{3\sqrt{5}} = \frac{4\sqrt{5}}{3}$
BETA
