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The graph of the equation $x^2 - xy - 2y^2 = 0$ is...

B) 2 intersecting lines.

I chose this answer following my 'gut feelings' on a MCQ, and it turned out to be right, but when I try to prove myself it's actually quite difficult and I get mixed up... Can someone give me a hint?

space
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3 Answers3

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Hint:

$$x^2-xy-2y^2=x(x-2y)+y(x-2y)=?$$

lab bhattacharjee
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You are correct, it can be rewritten in the following form $$(x+y)(x-2y)=0$$ so you have a pair of lines $y=-x$ and $2y=x$

Harry Alli
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It’s clearly a degenerate conic of some sort. There are no linear terms, so it’s centered at the origin, and it also includes the origin as a solution. The discriminant is $B^2-4AC=(-1)^2-4(1)(-2)=9\gt0$, so it’s a degenerate hyperbola, i.e., a pair of intersecting lines.

amd
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  • I'm curious because hyperbola graphs (or the two pairs of it) don't seem to intersect - so why are they a pair of intersecting lines? When I drew the graph, I got a hyperbola as well – space Nov 29 '17 at 00:08
  • @Helena No, it really is a pair of intersecting lines, so it seems you’ve made a mistake in plotting the graph. The key word here is degenerate. In fact, these lines are the asymptotes of the family of hyperbolas $x^2-xy-2y^2=c$. It might be helpful to think of these curves as horizontal slices through the saddle surface $z=x^2-xy-2y^2$; when the slice passes through the saddle point at the origin, you get a pair of intersecting lines. – amd Nov 29 '17 at 00:29
  • @Helena You can find an illustration of such a family of hyperbolas here. You can, of course, simply factor the equation to see that it represents a pair of lines, as is done in other answers. – amd Nov 29 '17 at 00:35