Question

Find the foci and the eccentricity of the conics $x^2+4xy+y^2-2x+2y-6=0.$

Answer 1

Differentiating partially w.r.t. x and y we get,
$\frac{\partial f}{\partial x}=x+2y-1=0$ and$\frac{\partial f}{\partial x}=2x+y+1=0$
Hence, $x=-1,y=1$
$\therefore$ The center is $(-1,1)$, and
$c^{{}^{\prime }}=gx+fy+c$
$=1+1-6=-4$
The equation referred to parallel axes through the center will be
$x^2+2xy+y^2=4$
Here, $\tan 2\theta =\frac{2h}{a-b}=\infty$
$\therefore 2\theta =90°$ or $270°$
$\theta =45°$ or $135°$
$\tan \theta =\pm 1$
$\cos \theta =\frac{1}{\sqrt{2}},\sin \theta =\frac{1}{\sqrt{2}}$
Now, $r^2=\frac{4(1+{\tan }^2\theta )}{1+4\tan \theta +{\tan }^2\theta }=\frac{4}{3}$ or $-4$
$\sqrt{r_1^2-r_2^2}=\frac{4}{3}\sqrt{3}$
So the co ordinates of the foci are
$\{x\pm \sqrt{r_1^2-r_2^2}\cos \theta ,y\pm \sqrt{r_1^2-r_2^2}\sin \theta \}$
i.e., $(-1\pm \frac{2}{3}\sqrt{6},1+\frac{2}{3}\sqrt{6})$
and $e^2=\frac{{\alpha }^2-{\beta }^2}{{\alpha }^2}=\frac{4+\frac{4}{3}}{\frac{4}{3}}=4$
$\therefore e=2$