The area of the region bounded by the ellipse x 2 25 + y 2 16 = 1 is:

Question

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

Smart Achievers · Accepted Answer

(1)

Ans. $20 π$ sq. units

Explanation:

We have

$\frac{x^{2}}{5^{2}} + \frac{y^{2}}{4^{2}} = 1$

Here, $a = \pm 5$ and $b = \pm 4$

and $\frac{y^{2}}{4^{2}} = 1 - \frac{x^{2}}{5^{2}}$

$\Rightarrow$ $y^{2} = 16 (1 - \frac{x^{2}}{25})$

$\Rightarrow$ $y = \sqrt{\frac{16}{25} (25 - x^{2})}$

$\Rightarrow$ $y = \frac{4}{5} \sqrt{5^{2} - x^{2}}$

Therefore, the area enclosed by the ellipse is

$A = 2 \cdot \frac{4}{5} \int_{- 5}^{5} \sqrt{5^{2} - x^{2}} d x$

$= 2 \cdot \frac{8}{5} \int_{0}^{5} \sqrt{5^{2} - x^{2}} d x$

$= 2 \cdot \frac{8}{5} {[\frac{x}{2} \sqrt{5^{2} - x^{2}} + \frac{5^{2}}{2} \sin^{- 1} (\frac{x}{5})]}_{0}^{5}$

$= 2 \cdot \frac{8}{5} [\frac{5}{2} \sqrt{5^{2} - 5^{2}} + \frac{5^{2}}{2} \sin^{- 1} (\frac{5}{5}) - 0 - \frac{25}{2} \cdot 0]$

$= 2 \cdot \frac{8}{5} [\frac{25}{2} \cdot \frac{π}{2}]$

$= \frac{16}{5} \cdot \frac{25 π}{4}$

$= 20 π sq. units.$

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