Question

Find the products of the semi-axes of the conics $y^2-4xy+5x^2=2.$

Answer 1

$5x^2-4xy+y^2=\mathrm{2.....}\left(i\right)a=5,b=1,h=-2$

$\tan 2\theta =\frac{2h}{a-b}$

$\tan 2\theta =\frac{2\times -2}{5-1}=-1$

$\frac{2\tan \theta }{1-{\tan }^2\theta }=-1{\tan }^2\theta -2\tan \theta -1=0$

$\tan \theta =\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}$

$\tan {\theta }_1=1+\sqrt{2},\tan {\theta }_2=1-\sqrt{2}$

$x=r\cos \theta ,y=r\sin \theta$

$r^2(5{\cos }^2\theta -4\sin \theta \cos \theta +{\sin }^2\theta )=2({\sin }^2\theta +{\cos }^2\theta )$

$r^2=\frac{2({\sin }^2\theta +{\cos }^2\theta )}{5{\cos }^2\theta -4\sin \theta \cos \theta +{\sin }^2\theta }$

$r^2=\frac{2{\tan }^2\theta +2}{5-4\tan \theta +{\tan }^2\theta }$

$\tan {\theta }_1=1+\sqrt{2}{r_1}^2=\frac{{2(1+\sqrt{2})}^2+2}{5-4(1+\sqrt{2})+(1+\sqrt{2}{)}^2}=\frac{8+4\sqrt{2}}{4-2\sqrt{2}}$

$\Rightarrow r_1=\sqrt{\frac{8+4\sqrt{2}}{4-2\sqrt{2}}}$

$\tan {\theta }_2=1-\sqrt{2}$

${r_2}^2=\frac{{2(1-\sqrt{2})}^2+2}{5-4(1-\sqrt{2})+(1-\sqrt{2}{)}^2}=\frac{8-4\sqrt{2}}{4+2\sqrt{2}}{r}_2=\sqrt{\frac{8-4\sqrt{2}}{4+2\sqrt{2}}}$

Product of semi axes$=r_1{r}_2$

$r_1{r}_2=\sqrt{\frac{8+4\sqrt{2}}{4-2\sqrt{2}}}\times \sqrt{\frac{8-4\sqrt{2}}{4+2\sqrt{2}}}{r}_1{r}_2=\sqrt{\frac{2(4+2\sqrt{2)}}{4-2\sqrt{2}}}\times \sqrt{\frac{2(4-2\sqrt{2)}}{4+2\sqrt{2}}}{r}_1{r}_2=\sqrt{2}\times \sqrt{2}=2$