JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Parabola

Let

P (a, b)

be a point on the parabola

y^{2} = 8 x

such that the tangent at P passes through the centre of the circle

x^{2} + y^{2} - 10 x - 14 y + 65 = 0

. Let

A

be the product of all possible values of

a

and

B

be the product of all possible values of

b

. Then the value of

A + B

is equal to :

(A)0
(B)25
(C)40
(D)65✓ Correct

Answer: (D)

Solution

The parabola is

y^{2} = 8 x

with parameter

a = 2

. A point on it is

P (2 t^{2}, 4 t)

.

The tangent at

P

t y = x + 2 t^{2}

.

The circle

x^{2} + y^{2} - 10 x - 14 y + 65 = 0

has center

(5, 7)

.

The tangent passes through

(5, 7)

7 t = 5 + 2 t^{2}

2 t^{2} - 7 t + 5 = 0

(2 t - 5) (t - 1) = 0

t = 1

t = 5/2

For

t = 1

P (2, 4)

, so

a = 2, b = 4

.
For

t = 5/2

P (25/2, 10)

, so

a = 25/2, b = 10

A = 2 \times \frac{25}{2} = 25

B = 4 \times 10 = 40

A + B = 65

. Answer: (D)

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option

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