Correct option is (A) 20π sq. unit
Given equation of ellipse is \(\frac{x^2}{25} + \frac{y^2}{16} = 1\)
⇒ \(\frac{y^2}{16} = 1 - \frac {x^2}{25}\)
⇒ \(y^2 = \frac{16}{25} (25 - x^2)\)
∴ \(y = \frac 45 \sqrt{25 - x^2}\)
∴ Since the ellipse is symmetrical about the axes.
∴ Required area = \(4 \times \int \limits_0^5 \frac45\sqrt{25 - x^2}\,dx\)
\(= 4 \times \frac 45 \int \limits_ 0 ^5 \sqrt{(5)^2 - x^2}\,dx\)
\(= \frac {16}5 \left[\frac x2 \sqrt{(5)^2 - x^2} + \frac{25}2\sin^{-1}\frac x5\right]_0^5\)
\(= \frac {16}5 \left[0 + \frac{25}2. \sin^{-1} (\frac 55) - 0 - 0\right]\)
\(= \frac{16}5 \left[\frac{25}2 .\sin^{-1}(1)\right]\)
\(= \frac{16}5 \left[\frac{25}2 .\frac \pi 2\right]\)
= 20π sq. units
