# 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 $C^2 > 4AC$ as here, it's a hyperbola. Therefore I don't get how the equation can represent 2 straight lines. Any clues?

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2
Conceptually, simply recall that a conic section is, of course, the intersection of a plane with a cone. (See en.wikipedia.org/wiki/Conic_section ) When the plane happens to pass through the apex of the cone, you get the degenerate cases: the "point" ellipse/circle, the "line" parabola, and the "crossed lines" hyperbola (which is effectively its own set of asymptotes). – Blue Jan 21 at 17:41

## 1 Answer

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 $f(x,y) = a(x,y)b(x,y)$.

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Thanks very much! I think I need to brush up my long division :/ – Luigi Plinge Jan 21 at 17:40

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One trick for performing the division is to write $f(x,y) = (x+2y+3)(ax+by+c)$ and then equate coefficients to find $a,b,c$. – user7530 Jan 21 at 17:43
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