$2x^2 + xy + 2y^2 = 0$ a pair of straight lines Is the following equation a Pair of Straight lines ?
$2x^2 + xy + 2y^2 = 0$
I can see $h^2 - ab$  is negative. I do not think it will be a Pair of Straight lines. Then what is it ?
Can anyone please explain ?
 A: $x^2+y^2+x\frac y2=0\implies x^2+2x(\frac y4)+(\frac y4)^2+15y^2/16=0\implies (x+y/4)^2+15 y^2/16=0$, which is possible on reals if and only if $(x,y)=(0,0)$ and hence the equation represents a point (degenerate conic).
A: The polynomial is homogeneous of the second order. You can factor it by rewriting
$$2\frac{x^2}{y^2}+\frac xy+2=0.$$
Then solving the quadratic,
$$\frac xy=\frac{-1\pm i\sqrt{15}}4,$$ gives the complex straight lines
$$4x+(1-i\sqrt{15})y=0$$ and $$4x+(1+i\sqrt{15})y=0.$$
A: Plotting the graph of this function on desmos tells me that well its a...... Point!

In general to know whether it is a point check for $h^2 < ab$ and $ \Delta = 0$ where $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$
Pair of Straight Lines would satisfy $\Delta = 0$ and $h^2 = ab$
EDIT: In general theory the condition I wrote for a point is the condition for the conic to represent non real lines 
References: https://brilliant.org/wiki/conics-discriminant/ 
A: Using $|xy|\le \frac 12(x^2+y^2)\quad$  (from $(|x|-|y|)^2\ge 0$)
You get that $\, 2x^2+xy+2y^2\ge \frac 32(x^2+y^2)$ so this can be zero, only for $x=y=0$ and your locus is degenerated in the origin point only.
