Question

The conic represented by the equation $x^2+y^2+2xy+4x-6y+17=0$ is

• A. a hyperbola
• B. a parabola
• C. a pair of straight lines
• D. an ellipse

Answer 1

The correct option is B a parabola

$x^2+y^2+2xy+4x-6y+17=0$
On comparing the given equation with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, we get
$a=b=1,c=17,g=2,f=-3,h=1$
$\therefore \Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2=-25\ne 0$
and $h^2-ab=0$
Hence, $\Delta \ne 0$ and $h^2=ab$
So the given equation represents a parabola.