Mathematics · Area under the Curves

JEE Main 2025 — 4 April, Evening Shift — Question 29

A line passing through the point $A(-2,0)$ , touches the parabola $P: y^{2}=x-2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $A B$ , parabola $P$ and the $x$ -axis, is:

Option A:
$\frac{8}{3}$
Correct
Option B:
3
Option C:
2
Option D:
$\frac{7}{3}$

Answer: A

Step-by-step solution

$y^{2}=4\left(\frac{1}{4}\right)(x-2)$

$y=m(x-2)+\frac{1}{4 m}$ passes through $(-2,0)$

$\Rightarrow 0=-4 m+\frac{1}{4 m} \Rightarrow 16 m^{2}=1$

$\Rightarrow m= \pm \frac{1}{4}$

$m=\frac{1}{4}$ in first quadrant $\Rightarrow$ contact point $(6,2)$

$\Rightarrow$ Area $=\frac{1}{2} \times(1) \times 4+\int_{2}^{6}\left[\left(\frac{x+2}{4}\right)-\sqrt{x-2}\right] d x$

=2+\frac{2}{3}=\frac{8}{3}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves