Question

The area of the region 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) 7
(d) none of these

Answer 1

The tangent passes through the point with ordinate 3, so substituting y = 3 in equation of parabola (y − 2)² = x − 1, we get x = 2
Therefore, the line touches the parabola at (2, 3).
We have,
$$\begin{gathered} {\left(y-2\right)}^2=x-1 \\ \Rightarrow y-2=\sqrt{x-1} \\ \Rightarrow y=\sqrt{x-1}+2 \end{gathered}$$
Slope of the tangent of parabola at x = 2
${\left[\frac{\mathit{d}y}{\mathit{d}x}\right]}_{x=2}={\left[\frac{1}{2\sqrt{x-1}}\right]}_{x=2}=\frac{1}{2}$
Therefore, the equation of the tangent is given as:
$$\begin{gathered} y-y_0=m\left(x-x_0\right) \\ \Rightarrow y-3=\frac{1}{2}\left(x-2\right) \\ \Rightarrow y=\frac{1}{2}x+2 \end{gathered}$$
Therefore, area of the required region ABC,
$$\begin{gathered} A={\int }_0^3\left(x_1-x_2\right)\mathit{d}y\mathit{}\mathit{}\mathit{}\left[\mathrm{Where},x_1={\left(y-2\right)}^2+1\mathrm{and}{x}_2=2\left(y-2\right)\right] \\ ={\int }_0^3\left(x_1-x_2\right)dy \\ ={\int }_0^3{\left(y-2\right)}^2+1-2\left(y-2\right)dy \\ ={\int }_0^3{\left[\left(y-2\right)-1\right]}^{2}dy \\ ={\int }_0^3{\left[y-3\right]}^{2}dy \\ ={\left[\frac{{\left(y-3\right)}^3}{3}\right]}_0^3 \\ =\left[\frac{{\left(3-3\right)}^3}{3}\right]-\left[\frac{{\left(0-3\right)}^3}{3}\right] \\ =9 \end{gathered}$$

Therefore the answer is (d)