Question
Area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
✓
Answer
$(c) :$ Total area, $A=4 \times$ Area in first quadrant
$=4 \times \int_0^a y d x=4 \int_0^a \frac{b}{a} \sqrt{a^2-x^2} d x$
$=\frac{4 b}{a}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a$
$=\pi a b \text { sq. units }$
$=4 \times \int_0^a y d x=4 \int_0^a \frac{b}{a} \sqrt{a^2-x^2} d x$
$=\frac{4 b}{a}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a$
$=\pi a b \text { sq. units }$
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