The area enclosed by the curve y^2 = 4ax and its latus-rectum is Option 1: 8/3a^2 Option 2: 4/3a^2 Option 3: 1/3a^2 Option 4: 1/12a^2

Answer: 8/3a^2. Solution: Latus rectum of y^2=4ax is the line x=a. Area = 2_0^a √4ax dx = 4√a_0^a √x dx = 4√a * 2/3a^(3/2) = 8/3a^2.

Calculus

Application of Integrals

Medium

The area enclosed by the curve $y^2 = 4ax$ and its latus-rectum is

$\frac{8}{3}a^2$

$\frac{4}{3}a^2$

$\frac{1}{3}a^2$

$\frac{1}{12}a^2$

✅ Correct Option: 1

Latus rectum of $y^2=4ax$ is the line $x=a$ .

Area $= 2\int_0^a \sqrt{4ax}\,dx = 4\sqrt{a}\int_0^a \sqrt{x}\,dx = 4\sqrt{a} \cdot \frac{2}{3}a^{3/2} = \frac{8}{3}a^2$ .

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