The area of the region bounded by the ellipse x^$\frac{2}{25}$ + y^$\frac{2}{16}$ = 1 is ______.

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KollegeApply · Accepted Answer

The area of the region bounded by the ellipse x^$\frac{2}{25}$ + y^$\frac{2}{16}$ = 1 is 20$\pi$ sq.unit. Explanation: Given equation of ellipse is x^$\frac{2}{25}$ + y^$\frac{2}{16}$ = 1 $\Rightarrow$ y^$\frac{2}{16}$ = 1 - x^$\frac{2}{25}$ $\Rightarrow$ y2 = $\frac{16}{25}$ (25 - x^2) $\therefore$ y = $\frac{4}{5}$ sqrt(25 - x^2) $\therefore$ Since the ellipse is symmetrical about the axes. $\therefore$ Required area = `4 xx int_0^5 $\frac{4}{5}$ sqrt(25 - x^2) "d"x = 4 xx $\frac{4}{5}$ int_0^5 sqrt((5)^2 - x^2) "d"x = $\frac{6}{5}$ [x/2 sqrt((5)^2 - x^2) + $\frac{25}{2}$ sin^-1 x/5]_0^5 = $\frac{16}{5}$ [0 + $\frac{25}{2}$ * sin^-1 ($\frac{5}{5}$) - 0 - 0] = $\frac{16}{5}$ [$\frac{25}{2}$ * sin^-1 (1)] = $\frac{16}{5}$ [$\frac{25}{2}$ * pi/2]` = 20$\pi$ sq.unit [Solution diagram: 251754_solution_1.png]

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