The area enclosed by the curve y2 = 4ax

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Application of Integrals

Question:

The area enclosed by the curve y² = 4ax and its latus - rectum is

Options:

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

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

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

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

Correct Answer:

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

Explanation:

y² = 4ax and latus rectum

area of I = area of II

So, area = $2 × \int\limits_0^a y d x$

$=2 \times \int\limits_0^a 2 \sqrt{a} \sqrt{x} d x$

$=2 \times 2 \sqrt{a} \times \frac{2}{3}\left[x^{3 / 2}\right]_0^a$

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

Option: 1