Question

Trace the following central conics.
$xy-y^2=a^2.$

Answer 1

$xy-y^2=a^{2}xy-y^2-a^2=0a=0,b=-1,c=-a^2,h=\frac{1}{2},f=0,g=0\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2\Delta =0+0-0-0-(-a^2)\frac{1}{4}=\frac{a^2}{4}\Delta \ne 0h^2=\frac{1}{4},ab=0h^2>ab$

So the conic is a hyperbola

$\tan 2\theta =\frac{2h}{a-b}\tan 2\theta =\frac{1}{0-(-1)}=1\frac{2\tan \theta }{1-{\tan }^2\theta }=1{\tan }^2\theta +2\tan \theta -1=0\tan \theta =\frac{-2\pm \sqrt{4+4}}{2}=-1\pm \sqrt{2}$

So the axes of hyperbola are located at $\tan {\theta }_1=-1+\sqrt{2}$ and $\tan {\theta }_2=-1-\sqrt{2}$

converting the equation of hyperbola in polar form

$x=r\cos \theta ,y=r\sin \theta r^2(\sin \theta \cos \theta -{\sin }^2\theta )=a^2({\sin }^2\theta +{\cos }^2\theta )r^2=\frac{a^2({\sin }^2\theta +{\cos }^2\theta )}{\sin \theta \cos \theta -{\sin }^2\theta }{r}^2=\frac{a^2{\tan }^2\theta +a^2}{\tan \theta -{\tan }^2\theta }=a^2\frac{{\tan }^2\theta +1}{\tan \theta -{\tan }^2\theta }\tan {\theta }_2=-1+\sqrt{2}{r_1}^2=a^2\frac{{(-1+\sqrt{2})}^2+1}{-1+\sqrt{2}-{(-1+\sqrt{2})}^2}=a^2\frac{4-2\sqrt{2}}{-4+3\sqrt{2}}\Rightarrow r_1=a\sqrt{\frac{4-2\sqrt{2}}{-4+3\sqrt{2}}}{r_2}^2=a^2\frac{{(-1-\sqrt{2})}^2+1}{-1+\sqrt{2}-{(-1-\sqrt{2})}^2}=a^2\frac{4+2\sqrt{2}}{-4-\sqrt{2}}{r}_2=a\sqrt{\frac{4+2\sqrt{2}}{-4-\sqrt{2}}}$

So the magnitude of axes is given by $r_1$ and $r_2$