Question

Trace the following central conics.
$x^2-2xycosec2\alpha +y^2=a^2.$

Answer 1

$x^2-2xy\csc 2\alpha +y^2=a^{2}a=1,b=1,h=-\csc 2\alpha \tan 2\theta =\frac{2h}{a-b}\tan 2\theta =\frac{-2\csc 2\alpha }{1-1}=\infty \frac{2\tan \theta }{1-{\tan }^2\theta }=\infty \Rightarrow 1-{\tan }^2\theta =0\tan \theta =\pm 1\theta ={45}^{\circ },{135}^{\circ }$

is the required location of axes.

converting the equation in polar from

$x=r\cos \theta ,y=r\sin \theta r^2({\cos }^2\theta -2\sin \theta \cos \theta \csc 2\alpha +{\sin }^2\theta )=a^2({\cos }^2\theta +{\sin }^2\theta )r^2=a^2\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta -2\sin \theta \cos \theta \csc 2\alpha +{\sin }^2\theta }{r}^2=a^2\frac{1+{\tan }^2\theta }{1-2\tan \theta \csc 2\alpha +{\tan }^2\theta }\tan {\theta }_1=1{r_1}^2=a^2\frac{1+1}{1-2\csc 2\alpha +1}\Rightarrow r_1=a\sqrt{\frac{1}{1-\csc 2\alpha }}\tan {\theta }_2=-1{r_2}^2=a^2\frac{1+1}{1+2\csc 2\alpha +1}{r}_2=a\sqrt{\frac{1}{1+\csc 2\alpha }}$

The magnitude of semi axes is given by $r_1$ and $r_2$

Hence the central conic can be traced.