Question

Find the area enclosed by the ellipse x2/a2 + y2/b2 =1??​

Answer 1

Step-by-step explanation:

$area \: of \: eclipe = 4 \times \int \: ydx \\ = > \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \\ = > {y}^{2} = \frac{ {b}^{2} }{ {a}^{2} } ( {a}^{2} - {x}^{2} ) \\ = > y = \frac{b}{a} \sqrt{ ( {a}^{2} - {x}^{2} ) } \\ = > area = 4 \times \int_ {0}^{a} \: \frac{b}{a} \sqrt{ ( {a}^{2} - {x}^{2} ) } dx \\ = > \frac{4b}{a} ( \frac{x}{2} \sqrt{ ( {a}^{2} - {x}^{2} ) } + \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} )_ {0}^{a} \\ = > \frac{4b}{a} (0 + \frac{ {a}^{2} }{2} {sin}^{ - 1}1 - 0) \\ = > \frac{4b}{a} \times \frac{ {a}^{2} }{2} \times \frac{\pi}{2} \\ = > \pi \: ab$