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

$20\pi$ sq units

The correct answer is Option (1) → $20\pi$ sq units

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

Here,  $a = \pm 5$ and $b = \pm 4$

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

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

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

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

$∴\text{Area enclosed by ellipse} = 4 \cdot \frac{4}{5} \int_{0}^{5} \sqrt{5^2 - x^2} \, dx$

$= \frac{16}{5} \left[ \frac{x}{2} \sqrt{5^2 - x^2} + \frac{5^2}{2} \sin^{-1} \frac{x}{5} \right]_{0}^{5}$

$\left[ ∵\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} \right]$

$= \frac{16}{5} \left[ \frac{5}{2} \sqrt{5^2 - 5^2} + \frac{5^2}{2} \sin^{-1} \frac{5}{5} - 0 - \frac{25}{2} \cdot 0 \right]$

$= \frac{16}{5} \left[ \frac{25}{2} \cdot \frac{\pi}{2} \right]$

$= \frac{16}{5} \cdot \frac{25\pi}{4} = 20\pi \text{ sq. units}$