question_answer

Area of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is                 [Karnataka CET 1993]

A)            \[\pi \,ab\]sq. unit             

B)            \[\frac{1}{2}\pi \,ab\]sq. unit

C)            \[\frac{1}{4}\pi \,ab\]sq. unit

D)            None of these

Correct Answer: A

Solution :

           Since the given equation contains only even powers of x and only even powers of y, the curve is symmetrical about y-axis as well as x-axis.                    \ Whole area of given ellipse                    \[=4(\text{area }\,\text{of}\,BCO)=4\times \int_{0}^{a}{y\,dx=4\int_{0}^{a}{\frac{b}{a}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}\]                    \[=4ab\int_{0}^{\pi /2}{\left( \frac{1+\cos 2\theta }{2} \right)\,d\theta }\],  {Putting \[x=a\sin \theta \]}                    \[=2ab\left( \int_{0}^{\pi /2}{\,\,d\theta +\int_{0}^{\pi /2}{\,\,\,\cos 2\theta \,d\theta }} \right)\]                    \[=[\theta ]_{0}^{\pi /2}+\left[ \frac{\sin 2\theta }{2} \right]_{0}^{\pi /2}=\pi ab\]sq. unit.