Question

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

• A. $\frac{32}{3}$
• B. $\frac{40}{3}$
• C. 16
• D. 19

Answer 1

Answer: (C)

16

Answer 2

The correct answer is (C) : 16
c1 : y² = 8x
c2 : y² = 16(3 - x)

Solving c 1 and c 2
48 – 16 x = 8 x
x = 2
∴ y = ± 4
∴ Area of shaded region
= $2$ \int_{0}^{4} \left\\{ \left( 48 - \frac{y^2}{16} \right) - \left( \frac{y^2}{8} \right) \right\\} \,dy
= $\frac{1}{8}$ $[48y - y^3]_{0}^{4} = 16$