Q79 / 170 · Manipal (MET) 2015 · Mathematics
The area of the region bounded by the parabola $(y-2)^2=(x-1)$, the tangent to the parabola at the point $(2,3)$ and the $X$-axis, is
- A$3$
- B$6$
- Correct answerC$9$
- D$12$
Solution
Given equation of parabola is
$y^2-4 y-x+5=0$
Then, the equation of tangent at $(2,3)$ is
$3 y-2(y+3)-\frac{(x+2)}{2}+5=0$
$\Rightarrow \quad 2 y-x-4=0$
$\therefore$ Required area is given by $\begin{aligned} A & =\int_0^3\left(x_2-x_1\right) d y \\ & =\int_0^3\left[\left\{(y-2)^2+1\right\}-\{2 y-4\}\right] d y \\ & =\int_0^3\left(y^2-6 y+9\right) d y=\int_0^3(3-y)^2 d y \\ & =-\left[\frac{(3-y)^3}{3}\right]_0^3\end{aligned}$ $\therefore \quad A=9$
$\therefore$ Required area is given by $\begin{aligned} A & =\int_0^3\left(x_2-x_1\right) d y \\ & =\int_0^3\left[\left\{(y-2)^2+1\right\}-\{2 y-4\}\right] d y \\ & =\int_0^3\left(y^2-6 y+9\right) d y=\int_0^3(3-y)^2 d y \\ & =-\left[\frac{(3-y)^3}{3}\right]_0^3\end{aligned}$ $\therefore \quad A=9$
Hence, the correct answer is option: 'C'.