Mathematics · Parabola

JEE Main 2025 — 2 April, Morning Shift — Question 35

Let the focal chord $P Q$ of the parabola $y^{2}=4 x$ make an angle of $60^{\circ}$ 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 $(0, \alpha)$ , then $5 \alpha^{2}$ is equal to:

Option A:
$15$
Correct
Option B:
$25$
Option C:
$30$
Option D:
$20$

Answer: A

Step-by-step solution

$P T: t y=x+a t^{2}$

$P S=P T$

$M_{t}=\frac{1}{t}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

$t=\sqrt{3}$

$\alpha=a t=\sqrt{3} \quad(a=1)$

$\therefore \quad 5 \alpha^{2}=15$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Parabola
Topic: Condition of tangents & normal