Question

The area bounded by the curve \[ y^2 = 4a^2(x - 1) \] and the lines \[ x = 1, \quad y = 4a \] is

• A. \( \frac{16}{3} a \) sq units
• B. \( \frac{16}{3} a^2 \) sq units
• C. \( 16 a^2 \) sq units
• D. \( 4a^2 \) sq units

Hint:

When calculating areas bounded by curves, set up an integral using the limits of the relevant region and perform the integration carefully.

Answer 1

The Correct Option is A

The equation of the curve is \( y^2 = 4a^2(x - 1) \). Rearranging this equation, we get: \[ x = \frac{y^2}{4a^2} + 1 \] We are asked to find the area bounded by the curve and the lines \( x = 1 \) and \( y = 4a \). Step 1: Set up the integral for the area The area bounded by the curve and the lines is given by the integral of the difference between the x-values at the curve and the line \( x = 1 \), for \( y \) between 0 and \( 4a \). The area \( A \) is: \[ A = \int_0^{4a} \left( \frac{y^2}{4a^2} + 1 - 1 \right) \, dy = \int_0^{4a} \frac{y^2}{4a^2} \, dy \] Step 2: Perform the integration Now, integrate \( \frac{y^2}{4a^2} \) with respect to \( y \): \[ A = \frac{1}{4a^2} \int_0^{4a} y^2 \, dy = \frac{1}{4a^2} \left[ \frac{y^3}{3} \right]_0^{4a} \] Substitute the limits of integration: \[ A = \frac{1}{4a^2} \times \frac{(4a)^3}{3} = \frac{1}{4a^2} \times \frac{64a^3}{3} = \frac{16}{3} a \] Thus, the area bounded by the curve and the lines is \( \frac{16}{3} a \) square units. Thus, the correct answer is \( \boxed{\frac{16}{3} a} \).