Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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 '13 at 17:41
    
The condition is $B^2>4AC$ and not $C^2>4AC$ –  Gaurav Apr 8 at 18:22

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
Thanks very much! I think I need to brush up my long division :/ –  Luigi Plinge Jan 21 '13 at 17:40
1  
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 '13 at 17:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.