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 _________ .

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none of these 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, \left( y - 2 \right)^2 = x - 1 \Rightarrow y - 2 = $\sqrt$(x - 1) \Rightarrow y = $\sqrt$(x - 1) + 2 Slope of the tangent of parabola at x = 2 \left[ d y/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: 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 Therefore, area of the required region ABC, A = \int_0^3 \left( x_1 - x_2 \right) dy ...........\left[\text{Where, }x_1 = \left( y - 2 \right)^2 + 1\text{ and }x_2 = 2\left( y - 2 \right) \right] = \int_0^3 \left( x_1 - x_2 \right) d y = \int_0^3 \left( y - 2 \right)^2 + 1 - 2\left( y - 2 \right) d y = \int_0^3 \left[ \left( y - 2 \right) - 1 \right]^2 d y = \int_0^3 \left[ y - 3 \right]^2 d y = \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 [Solution diagram: 43994_solution_1.png]

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