Question

Find the position and magnitude of the axes of the conics $12x^2-12xy+7y^2=48.$

Answer 1

$12x^2-12xy+7y^2=\mathrm{48.....}\left(i\right)a=12,b=7,h=-6\tan 2\theta =\frac{2h}{a-b}\tan 2\theta =\frac{-12}{12-7}=-\frac{12}{5}\frac{2\tan \theta }{1-{\tan }^2\theta }=-\frac{12}{5}6{\tan }^2\theta -5\tan \theta -6=06{\tan }^2\theta -9\tan \theta +4\tan \theta -6=03\tan \theta (2\tan \theta -3)+2(2\tan \theta -3)=0(3\tan \theta +2)(2\tan \theta -3)=0$

$\tan {\theta }_1=-\frac{2}{3}$ and $\tan {\theta }_2=\frac{3}{2}$ are the required position of axes.

Changing $\left(i\right)$ into polar coordinates

$x=r\cos \theta ,y=r\sin \theta r^2(12{\cos }^2\theta -12\sin \theta \cos \theta +7{\sin }^2\theta )=48({\sin }^2\theta +{\cos }^2\theta )r^2=\frac{48({\sin }^2\theta +{\cos }^2\theta )}{12{\cos }^2\theta -12\sin \theta \cos \theta +7{\sin }^2\theta }{r}^2=\frac{48{\tan }^2\theta +48}{12-12\tan \theta +7{\tan }^2\theta }\tan {\theta }_1=-\frac{2}{3}{r_1}^2=\frac{48\times \frac{4}{9}+48}{12+12\times \frac{2}{3}+7\times \frac{4}{9}}=3\Rightarrow r_1=\sqrt{3}$

$\tan {\theta }_2=\frac{3}{2}{r_2}^2=\frac{48\times \frac{9}{4}+48}{12-12\times \frac{4}{2}+7\times \frac{9}{4}}=16r_2=4$