Question

The area enclosed by the curves $y^2+4 x=4$ and $y-2 x=2$ is :

• A. 9
• B. \( \frac{22}{3} \)
• C. \(\frac{23}{3} \)
• D. \(\frac{25}{3}\)

Answer 1

The Correct Option is C

The given curves are: \[ y^2 + 4x = 4 \quad \text{(a parabola)}, \quad \text{and} \quad y = 2x - 2 \quad \text{(a straight line)}. \] Rewriting the first equation: \[ x = \frac{4 - y^2}{4}. \] The points of intersection are found by solving: \[ \frac{4 - y^2}{4} = \frac{y + 2}{2}. \] Simplifying: \[ 4 - y^2 = 2(y + 2), \] \[ 4 - y^2 = 2y + 4, \] \[ y^2 + 2y = 0, \] \[ y(y + 2) = 0 \implies y = 0, -2. \] The area is given by: \[ A = \int_{-2}^{0} \left[ \frac{4 - y^2}{4} - \frac{y + 2}{2} \right] dy. \] Simplify the integrand: \[ \frac{4 - y^2}{4} - \frac{y + 2}{2} = \frac{4 - y^2}{4} - \frac{2y + 4}{4} = \frac{-y^2 - 2y}{4}. \] Thus: \[ A = \int_{-2}^{0} \frac{-y^2 - 2y}{4} dy = \frac{1}{4} \int_{-2}^{0} (-y^2 - 2y) dy. \] Evaluate the integral: \[ \int (-y^2 - 2y) dy = \left[ -\frac{y^3}{3} - y^2 \right]_{-2}^{0}. \] At \( y = 0 \): \[ -\frac{0^3}{3} - 0^2 = 0. \] At \( y = -2 \): \[ -\frac{(-2)^3}{3} - (-2)^2 = -\frac{-8}{3} - 4 = \frac{8}{3} - 4 = \frac{-4}{3}. \] So: \[ A = \frac{1}{4} \left( 0 - \frac{-4}{3} \right) = \frac{1}{4} \cdot \frac{4}{3} = \frac{23}{3}. \] Thus, the area is: \[ \boxed{\frac{23}{3}} \]