JEE PYQ: Ellipse Question 18
Question 18 - 2019 (08 Apr Shift 1)
If the tangents on the ellipse $4x^2 + y^2 = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^2$ is equal to:
(a) $\frac{128}{17}$ (b) $\frac{64}{17}$ (c) $\frac{4}{17}$ (d) $\frac{2}{17}$
Show Answer
Answer: (d) $\frac{2}{17}$
Solution
$4a^2 + b^2 = 8$ …(1)
Tangent at $(1,2)$: $y = -2x + 4$
Tangent at $(a,b)$: slope $= \frac{-4a}{b}$
Perpendicular: $\frac{-4a}{b} \times (-2) = -1 \Rightarrow b = 8a$ …(2)
From (1) & (2): $a^2 = \frac{2}{17}$
BETA
