Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$ meet the $y$-axis at the points A and B respectively. Let the circle C be drawn taking AB as a diameter and the line $\mathrm{x}=2 \sqrt{5}$ intersect C at the points P and Q . If the tangents at the points P and Q on the circle intersect at the point ( $\alpha, \beta$ ) , then $\alpha^{2}-\beta^{2}$ is equal to

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Ellipse

Let the tangent and normal at the point

(33, 1)

on the ellipse

\frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1

meet the

y

-axis at the points A and B respectively. Let the circle C be drawn taking AB as a diameter and the line

x = 25

intersect C at the points P and Q . If the tangents at the points P and Q on the circle intersect at the point (

α, β

) , then

α^{2} - β^{2}

is equal to

Answer: (B)

The tangent at

(33, 1)

\frac{3 3 x}{36} + \frac{y}{4} = 1

, meeting the

y

-axis at

A = (0, 4)

. The tangent slope is

- \frac{x _{0}}{9 y _{0}} = - \frac{1}{3}

, so the normal has slope

3

and equation

y - 1 = 3 (x - 33)

, meeting the

y

-axis at

B = (0, - 8)

. The circle on

A B

as diameter has centre

(0, - 2)

, radius

6

, and equation

x^{2} + y^{2} + 4 y - 32 = 0

. For the external point

(α, β)

the chord of contact is

α x + β y + 2 (y + β) - 32 = 0

; equating it with the line

x = 25

gives

β + 2 = 0

, so

β = - 2

, and

\frac{α}{1} = \frac{2 β - 32}{- 2 5}

, so

α = \frac{18}{5}

. Hence

α^{2} - β^{2} = \frac{324}{5} - 4 = \frac{304}{5}

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option