JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Parabola

Let

y^{2} = 12 x

be the parabola and

S

be its focus. Let

P Q

be a focal chord of the parabola such that

(S P) (S Q) = \frac{147}{4}

. Let

C

be the circle described taking

P Q

as a diameter. If the equation of a circle

C

64 x^{2} + 64 y^{2} - α x - 643 y = β

, then

β - α

is equal to

Answer: 1328

Solution

For

y^{2} = 12 x

4 a = 12 \Rightarrow a = 3

, focus

S = (3, 0)

. For a focal chord with parameters

t, - 1/ t

P = (3 t^{2}, 6 t)

Q = (\frac{3}{t ^{2}}, \frac{- 6}{t})

, and

S P = 3 (1 + t^{2}), S Q = 3 (1 + \frac{1}{t ^{2}})

S P \cdot S Q = 9 \cdot \frac{( 1 + t ^{2} ) ^{2}}{t ^{2}} = \frac{147}{4} \Rightarrow \frac{( 1 + t ^{2} ) ^{2}}{t ^{2}} = \frac{49}{12}

Setting

s = t^{2}

12 s^{2} - 25 s + 12 = 0 \Rightarrow s = \frac{4}{3}

\frac{3}{4}

.

Taking

t^{2} = \frac{3}{4}

with

t = - \frac{3}{2}

P = (\frac{9}{4}, - 33)

, and with

1/ t^{2} = \frac{4}{3}

t_{2} = 2/ 3

Q = (4, 43)

.

Circle on

P Q

as diameter:

(x - \frac{9}{4}) (x - 4) + (y + 33) (y - 43) = 0 \Rightarrow x^{2} + y^{2} - \frac{25}{4} x - 3 y - 27 = 0

Multiplying by

64

64 x^{2} + 64 y^{2} - 400 x - 643 y - 1728 = 0 \Rightarrow 64 x^{2} + 64 y^{2} - 400 x - 643 y = 1728

α = 400, β = 1728

, giving

β - α = 1328

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer