The area of the figure bounded by the parabola `(y-2)^2=x-1,` the tangent to it at the point with the ordinate `x=3,` and the `x-a xi s` is `7s qdotu n i t e s` (b) `6s qdotu n i t e s` `9s qdotu n i t e s` (d) None of these

Options

A. 7 sq. units
B. 6 sq. units
C. 9 sq. units
D. None of these

Answer

C

Explanation

Given parabola is `(y-2)^(2)=x-1`
`rArr" "(dy)/(dx)=(1)/(2(y-2))`
When `y=3, x=2`
`therefore" "(dy)/(dx)=(1)/(2(3-2))=(1)/(2).`
Tangent at `(2,3)" is "y-3 =(1)/(2)(x-2)or x-2y + 4 =0`

`therefore" Required area "=int_(0)^(3)((y-2)^(2)+1)dy-int_(0)^(3)(2y-4)dy`
`=|((y-2)^(3))/(3)+y|_(0)^(3)-|y^(2)-4y|_(0)^(3)`
`=(1)/(3)+3+(8)/(3)-(9-12)=9` sq. units.