Q. The area of the region enclosed by the curves $y=x^2-4x+4$ and $y^2=16-8x$ is :
A) $\frac83$
B) $\frac43$
C) 8
D) 5
Explanation
[Solution by Dr. Abubacker] With $t=x-2$: $t^4=-8t$ gives $t=0,-2$, so the curves meet at $(2,0)$ and $(0,4)$. Area $=\int_0^2\left(\sqrt{16-8x}-(x-2)^2\right)dx=\frac{16}3-\frac83=\frac83$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.