Question

Area of the region enclosed between the curves \(y^{2}=4(x+7)\) and \(y^{2}=5(2-x)\) is

• A. \(\frac{32\sqrt{2}}{3}\)
• B. \(\frac{8}{3}\)
• C. \(\frac{1}{6}\)
• D. \(24\sqrt{5}\)

Hint:

When finding the area between two curves, it is often easier to integrate with respect to \(y\) if the curves are given in the form \(y^2 = f(x)\).

Answer 1

The Correct Option is D

We are given the curves \(y^2 = 4(x+7)\) and \(y^2 = 5(2-x)\). To find the area enclosed between these curves, we first need to find the points of intersection. Equating the expressions for \(y^2\), we have \[ 4(x+7) = 5(2-x) \] \[ 4x + 28 = 10 - 5x \] \[ 9x = -18 \] \[ x = -2 \] Substituting \(x = -2\) into \(y^2 = 4(x+7)\), we get \[ y^2 = 4(-2+7) = 4(5) = 20 \] \[ y = \pm 2\sqrt{5} \] The points of intersection are \((-2, 2\sqrt{5})\) and \((-2, -2\sqrt{5})\). The area enclosed between the curves is given by \[ A = \int_{-2\sqrt{5}}^{2\sqrt{5}} (x_2 - x_1) \, dy \] where \(x_1\) and \(x_2\) are the x-coordinates of the curves expressed in terms of \(y\). From \(y^2 = 4(x+7)\), we have \(x_1 = \frac{y^2}{4} - 7\). From \(y^2 = 5(2-x)\), we have \(x_2 = 2 - \frac{y^2}{5}\). Then \[ A = \int_{-2\sqrt{5}}^{2\sqrt{5}} \left( 2 - \frac{y^2}{5} - \left( \frac{y^2}{4} - 7 \right) \right) \, dy \] \[ = \int_{-2\sqrt{5}}^{2\sqrt{5}} \left( 9 - \frac{y^2}{5} - \frac{y^2}{4} \right) \, dy \] \[ = \int_{-2\sqrt{5}}^{2\sqrt{5}} \left( 9 - \frac{9y^2}{20} \right) \, dy \] \[ = \left[ 9y - \frac{3y^3}{20} \right]_{-2\sqrt{5}}^{2\sqrt{5}} \] \[ = \left( 18\sqrt{5} - \frac{3(2\sqrt{5})^3}{20} \right) - \left( -18\sqrt{5} - \frac{3(-2\sqrt{5})^3}{20} \right) \] \[ = 36\sqrt{5} - \frac{6(2\sqrt{5})^3}{20} \] \[ = 36\sqrt{5} - \frac{6(40\sqrt{5})}{20} \] \[ = 36\sqrt{5} - 12\sqrt{5} \] \[ = 24\sqrt{5} \] Therefore, the area enclosed between the curves is \(24\sqrt{5}\).