Mathematics · Parabola

JEE Main 2025 — 23 January, Evening Shift — Question 13

Let the shortest distance from $(a, 0), a>0$ , to the parabola $y^{2}=4 x$ be 4 . Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:

Option A:
$x^{2}+y^{2}-6 x+5=0$
Correct
Option B:
$x^{2}+y^{2}-4 x+3=0$
Option C:
$x^{2}+y^{2}-10 x+9=0$
Option D:
$x^{2}+y^{2}-8 x+7=0$

Answer: A

Step-by-step solution

Normal at P

$y+t x=2 t+t^{3}$

$\uparrow$

$(a, 0)$

at $=2 t+t^{3}$

$\mathrm{a}=2+\mathrm{t}^{2}$

$\mathbb{R}\left(2+\mathrm{t}^{2}, 0\right)$

$\mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^{2}=16$

4 t^{2}=12 \Rightarrow t^{2}=3

$a=5 \quad \mathbb{R}(5,0)$

Focus (1, 0)

$(1,0) \&(5,0)$ will be tha end pts. of diameter

$\Rightarrow \mathrm{Eg}^{\mathrm{n}}$ of circle is

\begin{aligned} & (x-1)(x-5)+y^{2}=0 \\& x^{2}+y^{2}-6 x+5=0 \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Mathematics
Chapter: Parabola
Topic: Shortest distance between a Parabola and a Point/Line/Curve