JEE PYQ: Ellipse Question 16
Question 16 - 2020 (09 Jan Shift 1)
If $e_1$ and $e_2$ are the eccentricities of the ellipse, $\frac{x^2}{18} + \frac{y^2}{4} = 1$ and the hyperbola, $\frac{x^2}{9} - \frac{y^2}{4} = 1$ respectively and $(e_1, e_2)$ is a point on the ellipse, $15x^2 + 3y^2 = k$, then $k$ is equal to:
(a) 16 (b) 17 (c) 15 (d) 14
Show Answer
Answer: (a) 16
Solution
$e_1 = \frac{\sqrt{7}}{3}$, $e_2 = \frac{\sqrt{13}}{3}$
$k = 15 \cdot \frac{7}{9} + 3 \cdot \frac{13}{9} = \frac{105 + 39}{9} = 16$
BETA
