Question

The area enclosed by the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in sq. m, is.

• A. $\pi ab$
• B. $\frac{\pi }{4}ab$
• C. $\frac{\pi }{2}ab$
• D. None of these

Answer 1

Answer: (A)

$\pi ab$

The parametric equations of the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $x=a\cos t$, $y=a\sin t$
The given curve is an ellipse which is closed curve & is completely described while t varies 0 to $2\pi$
$\therefore$ $x\frac{dy}{dt}-y\frac{dx}{dt}=ab({\cos }^{2}t+{\sin }^{2}t)=ab$
$\therefore$ Required Area $=\frac{1}{2}{\int }_0^{2\pi }(x\frac{dy}{dt}-y\frac{dx}{dt})dt=\frac{1}{2}{\int }_0^{2\pi }abdt=\pi ab$