JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Ellipse

Let

θ

be the acute angle between the tangents to the ellipse

\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1

and the circle

x^{2} + y^{2} = 3

at their point of intersection in the first quadrant. The

tan θ

is equal to:

Answer: (C)

Solution

Find the intersection point in the first quadrant:

From

x^{2} + y^{2} = 3

and

\frac{x ^{2}}{9} + y^{2} = 1

:

Subtracting:

x^{2} - \frac{x ^{2}}{9} = 2

, so

\frac{8 x ^{2}}{9} = 2

, giving

x^{2} = \frac{9}{4}

x = \frac{3}{2}

.

Then

y^{2} = 3 - \frac{9}{4} = \frac{3}{4}

, so

y = \frac{3}{2}

.

For the ellipse

\frac{x ^{2}}{9} + y^{2} = 1

, differentiating:

\frac{2 x}{9} + 2 y \frac{d y}{d x} = 0

, so

\frac{d y}{d x} = - \frac{x}{9 y}

.

At

(\frac{3}{2}, \frac{3}{2})

m_{1} = - \frac{3/2}{9 3 /2} = - \frac{1}{3 3}

.

For the circle

x^{2} + y^{2} = 3

, differentiating:

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

, so

\frac{d y}{d x} = - \frac{x}{y}

.

At

(\frac{3}{2}, \frac{3}{2})

m_{2} = - \frac{3/2}{3 /2} = - 3

tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} = \frac{- \frac{1}{3 3} + 3}{1 + \frac{1}{3}} = \frac{\frac{8}{3 3}}{\frac{4}{3}} = \frac{2}{3}

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option