Parametrize a curve, Multivariable Calculus I am stuck on what seems to be an easy exercise. 
We have $f(x,y) = x^2 + 4xy + y^2 \mbox{ for all } (x,y)$ in $\mathbb{R}^2.$
Now we are supposed to find a parametrization of the intersection curve between $f(x,y)$ and $z = x + 3y.$ 
I've been stuck for hours now, anyone got any ideas/tips? 
Thank you!

Setting 'z= z' gives $x^2+ 4xy+ y^2= x+ 3y$.  The first thing I would do is "complete the square" on the left: $x^2+ 4xy+ 4y^2- 3y^2= (x+ 2y)- 3y^2= x+ 3y$.  Now, let u= x+ y and v= y then x+ 3y= u- v+ 3v= u+ 2v so the equation becomes $u^2- 3v^2= u+ 2v$ or $u^2- u- 3(v^2+ (2/3)v)= 0$ and complete the square again: $u^2- u+ 1/4- 3(v^2+ (2/3)v+ 1/9)= (u- 1/2)^2- 3(v+ 1/3)^2= 1/4+ 1/3= 7/12$.  Let p= u- 1/2 and q= v+ 1/3 so the equation becomes $p^2- 3q^2= 7/12$ or $\frac{p^2}{\frac{12}{7}}- \frac{q^2}{\frac{4}{7}}= 1$
