Question

Find the area of the region between the parabola x = 4y − y² and the line x = 2y − 3.

Answer 1

To find the point of intersection of the parabola x = 4y − y² and the line x = 2y − 3
Let us substitute x = 2y − 3 in the equation of the parabola.
\[2y - 3 = 4y - y^2 \]
\[ \Rightarrow y^2 - 2y - 3 = 0\]
\[ \Rightarrow \left( y + 1 \right)\left( y - 3 \right)\]
\[ \Rightarrow y = - 1, 3\]
Therefore, the points of intersection are D(−1, −5) and A(3, 3).
The area of the required region ABCDOA,
\[A = \int_{- 1}^3 \left| x_1 - x_2 \right| dy.......\text{ where, }x_1 = 4y - y^2\text{ and }x_2 = 2y - 3\]
\[ = \int_{- 1}^3 \left( x_1 - x_2 \right) dy .........\left( \because x_1 > x_2 \right)\]
\[ = \int_{- 1}^3 \left[ \left( 4y - y^2 \right) - \left( 2y - 3 \right) \right] d y\]
\[ = \int_{- 1}^3 \left( 4y - y^2 - 2y + 3 \right) d y\]
\[ = \int_{- 1}^3 \left( - y^2 + 2y + 3 \right) d y\]
\[ = \left[ - \frac{y^3}{3} + \frac{2 y^2}{2} + 3y \right]_{- 1}^3 \]
\[ = \left[ - \frac{y^3}{3} + y^2 + 3y \right]_{- 1}^3 \]
\[ = - \frac{3^3}{3} + 3^2 + 9 - \left( \frac{1}{3} + 1 - 3 \right)\]
\[ = - 3^2 + 3^2 + 9 - \frac{1}{3} - 1 + 3\]
\[ = 11 - \frac{1}{3}\]
\[ = \frac{32}{3}\text{ sq . units }\]