Question

The area of the smaller region bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$ and the line $2x+5y=10$ is

• A. $\frac{5\pi }{2}$ sq. units
• B. $\frac{\pi }{2}-1$ sq. units
• C. $\frac{5\pi }{2}-5$ sq. units
• D. $\frac{\pi }{2}$ sq. units

Answer 1

The correct option is C $\frac{5\pi }{2}-5$ sq. units


$\frac{x^2}{5^2}+\frac{y^2}{2^2}=1\text{and}\frac{x}{5}+\frac{y}{2}=1$
Points of intersection is $(0,2)$ and $(5,0)$
$\frac{x^2}{25}+\frac{y^2}{4}=1\Rightarrow y=\sqrt{4(\frac{25-x^2}{25})}=\frac{2}{5}\sqrt{25-x^2}$

Required area
$=\text{ar}\left(AFDEA\right)-\text{ar}\left(△ADE\right)$
$=\frac{2}{5}\underset{0}{\overset{5}{\int }}\sqrt{25-x^2}dx-\frac{1}{2}\times 5\times 2$
$=\frac{2}{5}{(\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5})}_0^5-5$
$=\frac{2}{5}(0+\frac{25}{2}\times \frac{\pi }{2}-0)-5$
$=\frac{5\pi }{2}-5$ sq. units