<p>The area bounded by the curve \(y^2=4ax\) and its latus rectum is: [MAU006]</p>
Step-by-Step Solution
Key Concept: Latus rectum: x = a. Area = 2\int_0^a \sqrt[2]{ax} dx = 4\sqrt{a} \int_0^a \sqrt{x} dx = 4\sqrt{a} \cdot [2/3 \cdot x^(3/2)]_0^a = 8a^2/3.
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<p>The parabola $y^2=4ax$ is symmetric about the x-axis. Latus rectum is at $x=a$.</p>
<p>Area = $2\int_0^a\sqrt{4ax}\,dx = 4\sqrt{a}\int_0^a\sqrt{x}\,dx = 4\sqrt{a}\cdot\left[\frac{2}{3}x^{3/2}\right]_0^a = 4\sqrt{a}\cdot\frac{2}{3}a^{3/2} = \frac{8a^2}{3}$</p>
<p>$$\boxed{A=\frac{8a^2}{3}\text{ sq. units}}$$</p>
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Correct Answer: D