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The area enclosed by the curve y2 = 4ax and its latus - rectum is |
$\frac{8}{3} a^2$ $\frac{4}{3} a^2$ $\frac{1}{3} a^2$ $\frac{1}{12} a^2$ |
$\frac{8}{3} a^2$ |
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 |
