JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 11)
A line passing through the point $\mathrm{A}(-2,0)$, touches the parabola $\mathrm{P}: y^2=x-2$ 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 :
3
$\frac{7}{3}$
$\frac{8}{3}$
2
Explanation
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$$\begin{aligned} & y^2=4\left(\frac{1}{4}\right)(x-2) \\ & y=m(x-2)+\frac{1}{4 m} \text { passes through }(-2,0) \\ & \Rightarrow 0=-4 m+\frac{1}{4 m} \Rightarrow 16 m^2=1 \\ & \Rightarrow m= \pm \frac{1}{4} \\ & m=\frac{1}{4} \text { in first quadrant } \Rightarrow \text { contact point }(6,2) \end{aligned}$$
$$ \begin{aligned} & \Rightarrow \text { Area }=\frac{1}{2} \times(1) \times 4+\int_2^6\left[\left(\frac{x+2}{4}\right)-\sqrt{x-2}\right] d x \\ & \quad=2+\frac{2}{3}=\frac{8}{3} \end{aligned}$$
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