JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Ellipse

If the tangents on the ellipse

4 x^{2} + y^{2} = 8

at the points

(1, 2)

and

(a, b)

are perpendicular to each other, then

a^{2}

is equal to:

Answer: (A)

Solution

Differentiate the ellipse

4 x^{2} + y^{2} = 8

implicitly:

8 x + 2 y \frac{d y}{d x} = 0

gives slope

- \frac{4 x}{y}

, so at

(1, 2)

the tangent slope is

- 2

. A perpendicular tangent has slope

\frac{1}{2}

, so at

(a, b)

we need

- \frac{4 a}{b} = \frac{1}{2}

, that is

b = - 8 a

. Requiring

(a, b)

on the ellipse,

4 a^{2} + 64 a^{2} = 8

, so

68 a^{2} = 8

and

a^{2} = \frac{2}{17}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option