JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Area Under Curves

The area of the region enclosed by the curves

y = x^{2} - 4 x + 4

and

y^{2} = 16 - 8 x

is :

Answer: (A)

Solution

Write

y = x^{2} - 4 x + 4 = (x - 2)^{2}

and shift

X = x - 2

, so the curves become

y = X^{2}

and

y^{2} = - 8 X

, two parabolas through the origin.

Solving simultaneously: with

X = - y

on the branch that meets the other curve for

y > 0

y^{2} = - 8 X = 8 y \Rightarrow y^{3/2} = 8 \Rightarrow y = 4, X = - 2,

along with the trivial intersection at the origin.

For

0 \leq y \leq 4

, the boundary from

y = X^{2}

X = - y

(further left) and from

y^{2} = - 8 X

X = - \frac{y ^{2}}{8}

(closer to the axis). The enclosed area is

\int_{0}^{4} (- \frac{y ^{2}}{8} - (- y)) d y = \int_{0}^{4} (y - \frac{y ^{2}}{8}) d y = [\frac{2}{3} y^{3/2} - \frac{y ^{3}}{24}]_{0}^{4} = \frac{2}{3} (8) - \frac{64}{24} = \frac{16}{3} - \frac{8}{3} = \frac{8}{3} .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option