The area of the region bounded by the curves `y^(2)=4a^(2)(x-1)` and the lines x = 1 and y = 4a, is

Options

A. `(21a)/(2)` sq unit
B. `(16)/(3)` sq unit
C. `(17a)/(3)` sq unit
D. `(16a)/(3)` sq unit

Answer

D

Explanation

We have,
`y^(2)=4a^(2)(x-1)`
`rArr(y-0)^(2)=4a^(2)(x-1)`
Clearly, this equation represents a parabola with vertex at (1, 0) as shown in figure. The area enclosed by `y^(2)=4a^(2)(x-1),x=1` and y=4a is the area of shaded portion in figure.

`=int_(0)^(4a)(x-1)dy`
`=int_(0)^(4a)(y^(2))/(4a^(2))dy" "[{:(becauseP(x,y)"lies on "y^(2)=4a^(2)(x-1)),(thereforex-1=y^(2)//4a^(2)):}]`
`=(1)/(4a^(2))[(y^(3))/(3)]_(0)^(4a)`
`=(1)/(4a^(2))((64a^(3))/(3))=(16a)/(3)` sq unit