Question

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

• A. $16\pi$
• B. $20\pi$
• C. $25\pi$
• D. $36\pi$

Answer 1

The correct option is B $20\pi$

Given,

$\frac{x^2}{16}+\frac{y^2}{25}=1$

take $y=0$

we get, $x=\pm 4$

ellipse cuts the x-axis at $(4,0),(-4,0)$

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

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

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

Area of ellipse $=2\times$ area of function $y=\pm \frac{5}{4}\sqrt{5^2-x^2}$

$A={\int }_{-4}^{4}ydx={\int }_{-4}^4\frac{5}{4}\sqrt{5^2-x^2}dx$

$=\left[\frac{5}{4}{\{\frac{x}{2}\sqrt{5^2-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5}\}}_{-4}^4\right]$

$=\frac{5}{4}[\{\frac{4}{2}\sqrt{5^2-4^2}+\frac{25}{2}{\sin }^{-1}\frac{4}{5}\}-\{\frac{-4}{2}\sqrt{5^2-(-4{)}^2}+\frac{25}{2}{\sin }^{-1}\frac{-4}{5}\}]$

Upon solving the above equation, we get,

$=\frac{5}{4}\left(40\frac{\pi }{5}\right)$

$=10\pi$

Area of ellipse $=2\times 10\pi =20\pi$