Question

Find the area containing by ellipse $2x^2+6xy+5y^2=1$

Answer 1

$2x^2+6xy+5y^2=1$ -ellipse

$2x^2+6xy+5y^2-1=0$

$5y^2+6xy+2x^2-1=0$

$y=\frac{-6x\pm \sqrt{36x^2-20(2x^2-1)}}{10}$

$y=\frac{-6x\pm \sqrt{20-4x^2}}{10}$

$y=\frac{-3x\pm \sqrt{5-x^2}}{5}$

When $y=0$

$\Rightarrow 3x=\pm \sqrt{5-x^2}$

$9x^2=5-x^2$

$10x^2=5$

$x=\pm \frac{1}{\sqrt{2}}$

Area $=\left|{\int }_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{-3x+\sqrt{5-x^2}}{5}dx\right|+\left|{\int }_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{-3x-\sqrt{5-x^2}}{5}dx\right|$

Area $=\left|{[\frac{-3x^2}{10}+\frac{x\sqrt{5-x^2}}{10}+\frac{5}{10}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)]}_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\right|+\left|{[\frac{-3x^2}{10}-\frac{x\sqrt{5-x^2}}{10}-\frac{5}{10}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)]}_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\right|$

Area $=|\frac{3}{10}+\frac{5}{10}({\sin }^{-1}\left(\frac{1}{\sqrt{2}\sqrt{5}}\right)-\sin \left(\frac{-1}{\sqrt{2}\sqrt{5}}\right)]|$ $+|\frac{-3}{10}-\frac{5}{10}({\sin }^{-1}\left(\frac{1}{\sqrt{2}\sqrt{5}}\right)-{\sin }^{-1}\left(\frac{-1}{\sqrt{2}\sqrt{5}}\right)]|$

Area $=|\frac{3}{5}-2\pi |=2\pi -\frac{3}{5}$.