JEE Main 2025 Apr 4 Shift 2, Mathematics Q9: Area as Definite Integral

JEE Main2025Apr 4, Shift 2Mathematics

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 :-

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

Solution

Tangent through $A (- 2, 0)$ : $y = m (x + 2)$ .
Substituting into $y^{2} = x - 2$ :

$m^{2} (x + 2)^{2} = x - 2 \Rightarrow m^{2} x^{2} + (4 m^{2} - 1) x + (4 m^{2} + 2) = 0$ .

For tangency, discriminant $= 0$ :

$(4 m^{2} - 1)^{2} - 4 m^{2} (4 m^{2} + 2) = 0 \Rightarrow 16 m^{4} - 8 m^{2} + 1 - 16 m^{4} - 8 m^{2} = 0 \Rightarrow 16 m^{2} = 1$ .

Taking $m = \frac{1}{4}$ (first quadrant), the tangent is $y = \frac{1}{4} (x + 2)$ .
Substituting back gives point of tangency $B = (6, 2)$ .

Use $y$ as the integration variable. For $0 \leq y \leq 2$ , the parabola gives $x_{par} = y^{2} + 2$ and the line gives $x_{line} = 4 y - 2$ .

For these values of $y$ , the parabola lies to the right of the line, so the bounded region has area

$A = \int_{0}^{2} [(y^{2} + 2) - (4 y - 2)] d y = \int_{0}^{2} (y^{2} - 4 y + 4) d y$

$= [\frac{y ^{3}}{3} - 2 y^{2} + 4 y]_{0}^{2} = \frac{8}{3} - 8 + 8 = \frac{8}{3}$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q9: Area as Definite Integral

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