JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Ellipse

If a tangent to the ellipse

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

meets the tangents at the extremities of its major axis at

B

and

C

, then the circle with

B C

as diameter passes through the point :

Answer: (C)

Solution

The ellipse is

\frac{x ^{2}}{4} + y^{2} = 1

, and the tangents at the ends of the major axis are

x = 2

and

x = - 2

. The tangent at

(2 cos θ, sin θ)

\frac{x cos θ}{2} + y sin θ = 1

, which meets those two lines at

B = (2, \frac{1 - cos θ}{sin θ}), C = (- 2, \frac{1 + cos θ}{sin θ})

The circle on

B C

as diameter is

(x - 2) (x + 2) + (y - y_{B}) (y - y_{C}) = 0

. Here

y_{B} y_{C} = \frac{1 - cos ^{2} θ}{sin ^{2} θ} = 1

and

y_{B} + y_{C} = \frac{2}{sin θ}

, so the circle is

x^{2} + y^{2} - \frac{2 y}{sin θ} - 3 = 0

A point lying on it for every

θ

must kill the

θ

-dependent term, so

y = 0

and

x^{2} = 3

. The circle therefore always passes through

(3, 0)

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option