JEE PYQ: Ellipse Question 2
Question 2 - 2021 (17 Mar Shift 2)
Let $L$ be a tangent line to the parabola $y^2 = 4x - 20$ at $(6, 2)$. If $L$ is also a tangent to the ellipse $\frac{x^2}{2} + \frac{y^2}{b} = 1$, then the value of $b$ is equal to:
(a) 11 (b) 14 (c) 16 (d) 20
Show Answer
Answer: (b) 14
Solution
Tangent to parabola $2y = 2(x + 6) - 20$
$\Rightarrow y = x - 4$
Condition of tangency for ellipse: $16 = 2(1)^2 + b$
$\Rightarrow b = 14$