JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q7

MathematicsSingle correct+4 / -1Question 7 of 90

The area of the region bounded by $y^2=8x$ and $y^2=16(3-x)$ is equal to:

\dfrac{32}3

\dfrac{40}3

Answer and worked solution

Official answer: C

Meet. $8x=48-16x$ gives $x=2$ , $y=\pm4$ .

Area (in $y$ ): $\displaystyle\int_{-4}^4\left(\frac{48-y^2}{16}-\frac{y^2}8\right)dy=\int_{-4}^4\left(3-\frac{3y^2}{16}\right)dy=24-8=16$ .

(C)

Topics:Area between curves Parabola

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 7