JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Parabola

Let the focal chord PQ of the parabola

y^{2} = 4 x

make an angle of

6 0^{\circ}

with the positive x -axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y -axis at the point

(0, α)

, then

5 α^{2}

is equal to :

(A)15✓ Correct
(B)25
(C)30
(D)20

Answer: (A)

Solution

For the parabola

y^{2} = 4 x

, the focus is

S = (1, 0)

.

A point on the parabola is

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

. The focal chord through P has slope

tan 6 0^{\circ} = 3

\frac{2 t - 0}{t ^{2} - 1} = 3

2 t = 3 (t^{2} - 1)

Solving:

t = 3

, so

P = (3, 23)

.

The circle with diameter PS has equation:

(x - 1) (x - 3) + (y - 0) (y - 23) = 0

x^{2} - 4 x + 3 + y^{2} - 23 y = 0

x = 0

(y-axis):

3 + y^{2} - 23 y = 0

y^{2} - 23 y + 3 = 0

y = 3

α = 3

, and

5 α^{2} = 5 \cdot 3 = 15

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option

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