JEE PYQ: Ellipse Question 20
Question 20 - 2019 (10 Apr Shift 1)
If the line $x - 2y = 12$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $\left(3, \frac{-9}{2}\right)$, then the length of the latus rectum of the ellipse is:
(a) 9 (b) $12\sqrt{2}$ (c) 5 (d) $8\sqrt{3}$
Show Answer
Answer: (a) 9
Solution
Tangent at $\left(3, \frac{-9}{2}\right)$: $\frac{3x}{a^2} - \frac{9y}{2b^2} = 1$
Comparing with $x - 2y = 12$:
$a^2 = 36$, $b^2 = 27$
Latus rectum $= \frac{2b^2}{a} = \frac{54}{6} = 9$
BETA
