Question

Find the area of the region bounded by the parabola $$ y^2 = 4ax \text{ and its latus rectum.} $$ 

• A. \( \frac{a^2}{2} \)
• B. \( a^2 \)
• C. \( \frac{a^2}{4} \)
• D. \( \frac{a^2}{8} \)

Hint:

To find the area under a curve, integrate the function over the desired limits. For curves like parabolas, carefully handle the limits and the integrals.

Answer 1

The Correct Option is A

Step 1: The standard equation for a parabola is \( y^2 = 4ax \). The latus rectum is defined as the line segment perpendicular to the axis of symmetry, passing through the focus, and connecting two points on the parabola. For this parabola, the focus is located at \( (a, 0) \), and the latus rectum is the vertical line \( x = a \).

Step 2: To determine the area, we will integrate the region enclosed by the parabola and its latus rectum. From the parabola's equation, we derive \( y = \pm \sqrt{4ax} \). The area can be computed using the following integral: \[ \text{Area} = 2 \int_0^a \sqrt{4ax} \, dx \]

Step 3: Simplifying the integral yields: \[ \text{Area} = 2 \int_0^a \sqrt{4a} \sqrt{x} \, dx = 2 \sqrt{4a} \int_0^a \sqrt{x} \, dx \] \[ = 2 \sqrt{4a} \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^a = 2 \sqrt{4a} \cdot \frac{2}{3} a^{3/2} \] \[ = \frac{4}{3} a^{2} \] Therefore, the area is \( \frac{a^2}{2} \).