JEE Main 2025 Apr 2 Shift 1, Mathematics Q20: Parabola: Chord, Tangent and Normal

JEE Main2025Apr 2, Shift 1Mathematics

Let the focal chord $P Q$ of the parabola $y^{2} = 4 x$ make an angle of $60°$ with the positive x-axis, where $P$ lies in the first quadrant. If the circle, whose one diameter is $P S$ , $S$ being the focus of the parabola, touches the y-axis at the point $(0, α)$ , then $5 α^{2}$ is equal to :

A
15
B
25
C
30
D
20

Solution

Find P. Parametrise the parabola $y^{2} = 4 x$ as $(t^{2}, 2 t)$ , with focus $S (1, 0)$ . The slope of $S P$ for parameter $t$ is:

$\frac{2 t - 0}{t ^{2} - 1} = tan 60° = 3$

For $P$ in the first quadrant take $t > 0$ . Solving $2 t = 3 (t^{2} - 1)$ gives $3 t^{2} - 2 t - 3 = 0$ , whose positive root is $t = 3$ . So $P = (3, 23)$ .

Circle with diameter PS. Endpoints $P (3, 23)$ and $S (1, 0)$ . The equation is:

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

Tangency to y-axis: set $x = 0$ :

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

$3 + y^{2} - 23 y = 0 \Rightarrow (y - 3)^{2} = 0$

Double root at $y = 3$ , confirming tangency. So $α = 3$ and:

$5 α^{2} = 5 \times 3 = 15$

$∴$ Option (1) is correct.

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JEE Main 2025 Apr 2 Shift 1, Mathematics Q20: Parabola: Chord, Tangent and Normal

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