Question

Find the products of the semi-axes of the conics $11x^2+16xy-y^2-70x-40y+82=0.$

Answer 1

$11x^2+16xy-y^2-70x-40y+82=0$

$\Delta \ne 0h^2>ab$

So the equation represents a hyperbola

$11x^2+16xy-y^2=70x+40y-82\frac{\partial }{\partial x}(11x^2+16xy-y^2-70x-40y+82=0)22x+16y-70=\mathrm{0........}\left(i\right)\frac{\partial }{\partial y}(11x^2+16xy-y^2-70x-40y+82=0)16x-2y-40=\mathrm{0......}\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$ we get the centre of the conic as origin

$x=\frac{13}{5},y=\frac{4}{5}C:(\frac{13}{5},\frac{4}{5})$

Making the conic central by using the center

$11x^2+16xy-y^2=cc=70x+40y-82c=70.\frac{13}{5}+40.\frac{4}{5}-82=13211x^2+16xy-y^2=\mathrm{132.......}\left(iii\right)a=11,b=-1,h=8\tan 2\theta =\frac{2h}{a-b}\tan 2\theta =\frac{2\times 8}{11+1}=\frac{4}{3}\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{4}{3}4{\tan }^2\theta +6\tan \theta -4=04{\tan }^2\theta +8\tan \theta -2\tan \theta -4=04\tan \theta (\tan \theta +2)-2(\tan \theta +2)=0(4\tan \theta -2)(\tan \theta +2)=0\tan {\theta }_1=\frac{1}{2},\tan {\theta }_2=-2$

Converting $\left(iii\right)$ in polar form

$x=r\cos \theta ,y=r\sin \theta r^2(11{\cos }^2\theta +16\sin \theta \cos \theta -{\sin }^2\theta )=132({\sin }^2\theta +{\cos }^2\theta )r^2=\frac{132({\sin }^2\theta +{\cos }^2\theta )}{11{\cos }^2\theta +16\sin \theta \cos \theta -{\sin }^2\theta }{r}^2=\frac{132{\tan }^2\theta +132}{11+16\tan \theta -{\tan }^2\theta }\tan {\theta }_1=\frac{1}{2}{r_1}^2=\frac{132.\frac{1}{4}+132}{11+16.\frac{1}{2}-\frac{1}{4}}=\frac{132.\frac{5}{4}}{\frac{75}{4}}=\frac{132}{15}=\frac{44}{5}\Rightarrow r_1=\sqrt{\frac{44}{5}}{r_2}^2=\frac{132\times 4+132}{11-32-4}=\frac{-660}{25}=\frac{-132}{5}{r}_2=\sqrt{\frac{-132}{5}}$

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

$r_1{r}_2=\sqrt{\frac{44}{5}}\sqrt{\frac{-132}{5}}=\frac{44\sqrt{-3}}{5}$