Question

Area in the first quadrant between the ellipses $x^2+2y^2=a^2$ and $2x^2+y^2=a^2$ is

• A. $\frac{a^2}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}$
• B. $\frac{3a^2}{4}{\tan }^{-1}\frac{1}{2}$
• C. $\frac{5a^2}{2}{\sin }^{-1}\frac{1}{2}$
• D. $\frac{9\pi a^2}{2}$

Answer 1

The correct option is A $\frac{a^2}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}$


Point of intersection of $x^2+2y^2=a^2$ and $2x^2+y^2=a^2$ is $(\frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}})$

Required area is,
$A=\frac{1}{\sqrt{2}}\underset{0}{\overset{a/\sqrt{3}}{\int }}\sqrt{a^2-x^2}dx+\underset{a/\sqrt{3}}{\overset{a/\sqrt{2}}{\int }}\sqrt{a^2-2x^2}dx$
$=\frac{1}{\sqrt{2}}{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_0^{a/\sqrt{3}}+{[\frac{x}{2}\sqrt{a^2-2x^2}+\frac{a^2}{2\sqrt{2}}{\sin }^{-1}\frac{\sqrt{2}x}{a}]}_{a/\sqrt{3}}^{a/\sqrt{2}}$
$=\frac{a^2}{\sqrt{2}}{\tan }^{-1}\frac{1}{\sqrt{2}}$