MCQ
Area of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is
- ✓$\pi \,ab\,\,\, sq. \,unit$
- B$\frac{1}{2}\pi \,ab\,\,\, sq. \,unit$
- C$\frac{1}{4}\pi \,ab\,\,\, sq. \,unit$
- DNone of these
✓
Answer
Correct option: A.
$\pi \,ab\,\,\, sq. \,unit$
a
(a) Since the given equation contains only even powers of $x$ and only even powers of $y$,
(a) 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.
$\therefore$ Whole area of given ellipse
$ = 4({\rm{area\,\, }}\,{\rm{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$
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