Question

Find the position and magnitude of the axes of the conics $3x^2+2xy+3y^2=8.$

Answer 1

$3x^2+2xy+3y^2=\mathrm{8.....}\left(i\right)a=3,b=3,h=1\tan 2\theta =\frac{2h}{a-b}\tan 2\theta =\frac{2}{3-3}=\infty \frac{2\tan \theta }{1-{\tan }^2\theta }=\infty \Rightarrow 1-{\tan }^2\theta =0\tan \theta =\pm 1\Rightarrow \theta ={45}^{\circ },{135}^{\circ }$

is the required position of axes.

Chnaging $\left(i\right)$ to plar coordinates

$x=r\cos \theta ,y=r\sin \theta r^2(3{\cos }^2\theta +2\sin \theta \cos \theta +3{\sin }^2\theta )=8({\sin }^2\theta +{\cos }^2\theta )r^2=\frac{8({\sin }^2\theta +{\cos }^2\theta )}{3{\cos }^2\theta +2\sin \theta \cos \theta +3{\sin }^2\theta }{r}^2=\frac{8{\tan }^2\theta +8}{3+2\tan \theta +3{\tan }^2\theta }\tan {\theta }_1=1{r_1}^2=\frac{8+8}{3+2+3}=2\Rightarrow r_1=\sqrt{2}\tan {\theta }_2=-1{r_2}^2=\frac{8+8}{3-2+3}=4r_2=2$