What is $x^2+4xy+y^2=1$ an equation of?? In an an exam i'm asked what is $x^2+4xy+y^2=1$ an equation of?? I haven't seen an equation of this form before so how do i work what it is even if i hadn't seen it before?
 A: There’s a quick way of making a qualitative determination: ignore the terms of degree less than $2$, to get something that looks like $Ax^2+Bxy+Cy^2$. Then look at the discriminant, $B^2-4AC\,$. If positive, you have a hyperbola; if negative, an ellipse; if zero, a parabola. Test this out on all the cases you’re familiar with, including $xy=1$, just to reassure yourself. Then, if you’re enthusiastic, make a rotation of the coordinate axes and see that $B^2-4AC$ doesn’t change!
EDIT 1: Whoops, I should also have said that an equation like this can also turn out to be a pair of lines, or a single line counted twice. I’ll leave it to you to see what the story is there.
EDIT 2: In the case at hand, you can also complete a square:
$$
x^2+4xy+y^2=(x+2y)^2-3y^2=1\,,
$$
so that your curve is a “scrunched” version of $X^3-3Y^2=1$: the transformation between $(x,y)$ and $(X,Y)$ is definitely not distance-preserving, but it’s enough to conclude that your curve is a hyperbola. The method of rotating and then examining is sound, but very slow and computation-intensive.
A: A clever way would be to observe that that $x^{2}+4xy+y^{2}=1 \iff (x+y)^{2} +2xy =1$ and also $x^{2}+4xy+y^{2} \iff (x-y)^{2}+6xy=1$ and hence we may get rid of the term $xy$  in this way, $$3\{(x+y)^{2}+2xy\}-\{(x-y)^{2}+6xy\}=3-1$$ and hence $$3(x+y)^{2}-(x-y)^{2}=2.$$ This is the equation of a herperbola.
