Question

The area of the figure bounded by the parabola ${(y-2)}^2=x-1$, the tangent to it at the point with the ordinate $3$ and the $x$-axis is

• A. $3$
• B. $6$
• C. $9$
• D. None of these

Answer 1

The correct option is C $9$

Given parabola, is
${(y-2)}^2=x-1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2(y-2)}$
when, $y=3,x=2$
$\frac{dy}{dx}=\frac{1}{2(3-2)}=\frac{1}{2}$
Tangent at $(x,3)$ is
$y-3=\frac{1}{2}(x-2)$
$\Rightarrow x-2y+4=0$
Therefore, required area is
${\int }_0^3\{{(y-2)}^2+1\}dy-{\int }_0^3(2y-4)dy$
$={[\frac{{(y-2)}^3}{3}+y]}_0^3-{[y^2-4y]}_0^3$
$=\frac{1}{3}+3+\frac{8}{3}-(9-12)=9$.