JEE PYQ: Ellipse Question 14
Question 14 - 2020 (07 Jan Shift 2)
If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$ for some $a \in R$, then the distance between the foci of the ellipse is:
(a) $2\sqrt{7}$ (b) 4 (c) $2\sqrt{5}$ (d) $2\sqrt{2}$
Show Answer
Answer: (a) $2\sqrt{7}$
Solution
$y = -\frac{3}{4}x + 3\sqrt{2}$
Condition of tangency: $18 = a^2 \cdot \frac{9}{16} + 9 \Rightarrow a^2 = 16$
$e = \frac{\sqrt{7}}{4}$, distance between foci $= 2\sqrt{7}$
BETA
