Question

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. $6\mathrm{sq}$ units
• B. $9\mathrm{sq}$ units
• C. $12\mathrm{sq}$ units
• D. $3\mathrm{sq}$ units

Answer 1

The correct option is B

$9\mathrm{sq}$ units

Explanation for the correct option:

Given, equation of the parabola,

Step-1 Equation of tangent on parabola:

$$\begin{gathered} \Rightarrow {\left(y-2\right)}^2=\left(x-1\right) \\ {y}^2-4y-x+5=0 \end{gathered}$$

Plotting the curves,

As we know, the equation of tangent to the parabola $y^2=4ax$ at point $\left(x_1,y_1\right)$ is $y{y}_1=2a\left(x+x_1\right)$.

For tangent at a point, replace the terms of general equation in the form,

$x^2=x{x}_1,y^2=y{y}_1,x=\left(\frac{x_1+x}{2}\right),y=\left(\frac{y_1+y}{2}\right)$

Then, the equation of parabola at $(2,3)$ is given by,

$$\begin{gathered} 3y-2(y+3)-\frac{\left(x+2\right)}{2}+5=0\left(\text{Here},x_1=2,y_1=3\right) \\ 2y-x-4=0 \end{gathered}$$

Step-2 Required area:

Required area $A$ is given by,$$\begin{gathered} A={\int }_0^3\left(x_2-x_1\right)dx \\ ={\int }_0^3\left[\left\{{\left(y-2\right)}^2+1\right\}-\left\{2y-4\right\}\right]dy \\ ={\int }_0^3\left(y^2-6y+9\right)dy \\ ={{\left[\frac{y^3}{3}+\frac{\left(-6\right)y^2}{2}+9y\right]}_0}^3\left(\because \int x^{n}dx=\frac{x^{n+1}}{n+1}\right) \\ ={{\left[\frac{y^3}{3}-3y^2+9y\right]}^3}_0 \\ =\left[\left(3^2-3^3+27\right)-\left(0\right)\right] \\ =9 \end{gathered}$$

Therefore Area $=9sq.units$

Hence, option(B) is the correct answer.