Question

Trace the following central conics.
$y^2-2xy+2x^2+2x-2y=0.$

Answer 1

$y^2-2xy+2x^2+2x-2y=0a=2,b=1,h=-1,g=1,f=-1\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2\Delta =0+2-2-1=-1\Delta \ne 0h^2=1ab=2h^2<ab\frac{\partial }{\partial x}(y^2-2xy+2x^2+2x-2y=0)4x-2y+2=02x-y+1=\mathrm{0.......}\left(i\right)\frac{\partial }{\partial y}(y^2-2xy+2x^2+2x-2y=0)2y-2x-2=0$

$y-x-1=\mathrm{0......}\left(ii\right)$

$x=0,y=1$

$C:(0,1)$

$y^2-2xy+2x^2=2y-2x$

$c=2y-2x$

$c=2$

$y^2-2xy+2x^2=2$

$\tan 2\theta =\frac{2h}{a-b}$

$\tan 2\theta =\frac{-2}{2-1}=-2$

$\frac{2\tan \theta }{1-{\tan }^2\theta }=-2$

$2{\tan }^2\theta -2\tan \theta -2=0{\tan }^2\theta -\tan \theta -1=0$

$\tan \theta =\frac{1\pm \sqrt{1+4}}{2}\tan \theta =\frac{1\pm \sqrt{5}}{2}$

is the position of the axes of the ellipse.