Question

The area bounded by the ellipse $25x^2+16y^2=400$ is _____

• A. $16\pi$
• B. $25\pi$
• C. $20\pi$
• D. $40\pi$

Answer 1

The correct option is C $20\pi$

$25x^2+16y^2=400$

$⟹\frac{x^2}{16}+\frac{y^2}{25}=1$

Here $a=4$ and $b=5$

we know that ellipse is symmetrically divided into 4 parts so we need to just find the area for one part only

$A_1$ is the area of quadrant and A represent total area of ellipse

$y=5\sqrt{1-\frac{x^2}{16}}$

$y=\frac{5}{4}\sqrt{16-x^2}$

$A_1=\frac{5}{4}{\int }_0^4\left(\sqrt{16-x^2}\right)dx$

$=\frac{5}{4}{[\frac{x}{2}\sqrt{16-x^2}+8\times {\sin }^{-1}\frac{x}{4}]}_0^4$ ...... $\left(\text{applying integration formula for the}\sqrt{a^2-x^2}\text{form}\right)$

$=\frac{5}{4}(2\sqrt{16-16}+8\times {\sin }^{-1}1)$

$=\frac{5}{4}(8\times \frac{\pi }{2})$

$A_1=5\pi$

and, $A=4A_1$

$\therefore A=4\times 5\pi$

$A=20\pi$