Hyperbola is a pair of straight lines? I'm confused by this question:

If $f(x) = 2x^2 - 6y^2+xy+2x-17y-12=0$ is to represent a pair of
  straight lines, one of which has equation $x+2y+3=0$, what must be the
  equation of the other line? Verify that $f(x)=0$ does, indeed,
  represent a pair of straight lines.

Given the general form of a conic section $Ax^2+By^2+Cxy+Dx+Ey+F=0$ we know that if $C^2 > 4AB$ as here, it's a hyperbola. Therefore I don't get how the equation can represent 2 straight lines. Any clues?
 A: Dividing $f(x,y)$ through by the suggested $x+2y+3$ gives
$$f(x,y) = (x+2y+3)(2x-3y-4)=0.$$
The product is zero when either $x+2y+3=0$ or $2x-3y-4=0$, both of which are equations for lines.
You're right that $f$ is has positive discriminant, but it happens to be a reducible degenerate conic. Maybe the simplest example is $y^2-x^2=0$, which is clearly a pair of lines. Generally speaking, a conic section $f(x,y)=0$ will be degenerate any time you can factor $f(x,y) = a(x,y)b(x,y)$.
A: Divide the function $f(x, y)=2x^2-6y^2+xy+2x-17y-12$ by given factor $(x+2y+3)$ to get the second linear factor as $(2x-3y-4)$ 
Hence, the equation of second straight line: $2x-3y-4=0$  
For general case, a quadratic equation of two variables $x$ & $y$ is given as
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ 
The above equation will represent a pair of straight lines iff the discriminant ($\Delta$) is satisfied as follows 
$$\Delta=abc+2fgh-af^2-bg^2-ch^2=0$$ 
As per your question, we have
$a=2$, $h=\frac{1}{2}$, $b=-6$, $g=1$, $f=\frac{-17}{2}$ & $c=-12$
$$\Delta=(2)(-6)(-12)+2\left(\frac{-17}{2}\right)(1)\left(\frac{1}{2}\right)-(2)\left(\frac{-17}{2}\right)^2-(-6)(1)^2-(-12)\left(\frac{1}{2}\right)^2=0$$
Hence, the given equation:$2x^2-6y^2+xy+2x-17y-12=0$ represents a pair of straight lines.
