Question

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

• A. 09
• B. 9
• C. 9.0
• D. 9.00

Answer 1

Answer: (A), (B), (C), (D)

Given, equation of parabola is
$(y-2{)}^2=(x-1)$
$\Rightarrow y^2-4y-x+5=0\dots \left(1\right)$
Given that ordinate of tangent is $3$ , so it lies on the parabola, $(3-2{)}^2=(x-1)$
$\Rightarrow 1=x-1\Rightarrow x=2$
On differentiating equation $\left(1\right)$
$2y\frac{dy}{dx}-4\frac{dy}{dx}-1=0$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,3)}=\frac{1}{2y-4}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,3)}=\frac{1}{2}$
The equation of tangent at $(2,3)$
$\Rightarrow y-3=\frac{1}{2}(x-2)$
$\Rightarrow 2y-6=x-2$
$\Rightarrow 2y-x-4=0$



$\therefore$ Required area $A$ is given by
$A={\int }_0^3\left[\right\{(y-2{)}^2+1\}-\{2y-4\}]dy$
$A={\int }_0^3(y^2-6y+9)dy$
$\Rightarrow A={\int }_0^3(3-y{)}^{2}dy=-{\left[\frac{(3-y{)}^3}{3}\right]}_0^3$
$\therefore A=9$