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(B) The equation of the parabola is $(y-2)^2 = x-1$.The tangent at the point $(2,3)$ is found using the formula $y-y_1 = m(x-x_1)$.First,differentiate

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The equation of the parabola is $(y-2)^2 = x-1$.The tangent at the point $(2,3)$ is found using the formula $y-y_1 = m(x-x_1)$.First,differentiate the parabola equation with respect to $x$: $2(y-2) \frac{dy}{dx} = 1$,so $\frac{dy}{dx} = \frac{1}{2(y-2)}$.At $(2,3)$,the slope $m = \frac{1}{2(3-2)} = \frac{1}{2}$.The equation of the tangent is $y-3 = \frac{1}{2}(x-2)$,which simplifies to $2y-6 = x-2$,or $x = 2y-4$.The region is bounded by the parabola $x = (y-2)^2 + 1$,the tangent $x = 2y-4$,and the $x$-axis $(y=0)$.The intersection of the tangent with the $x$-axis occurs at $y=0$,which gives $x = 2(0)-4 = -4$.The intersection of the parabola with the $x$-axis occurs at $y=0$,which gives $x = (0-2)^2 + 1 = 5$.However,the region is bounded by the tangent and the parabola up to the point $(2,3)$. Integrating with respect to $y$ from $y=0$ to $y=3$:Area $A = \int_{0}^{3} [x_{parabola} - x_{tangent}] dy = \int_{0}^{3} [((y-2)^2 + 1) - (2y-4)] dy$$A = \int_{0}^{3} [y^2 - 4y + 4 + 1 - 2y + 4] dy = \int_{0}^{3} [y^2 - 6y + 9] dy$$A = \int_{0}^{3} (y-3)^2 dy = \left[ \frac{(y-3)^3}{3} \right]_{0}^{3} = 0 - \left( \frac{-27}{3} \right) = 9$.

Find 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.

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