Parametrize an intersection of a plane and an elliptic paraboloid I'm supposed to parametrize the intersection of the plane that has the equation $z = 5x + 3y$ and the 'elliptic paraboloid' with the equation $z = 3x^2+2xy+3y^2$
These two equations can also be written in terms of $u$ and $v$, the plane then has equation $z = 4u + v$ and the paraboloid has $z= 2u^2 + v^2$
How can I go about parametrising these in either domain? 
Thanks in advance

The intersection satisfies
$$ 4u + v = 2u^2 + v^2 \implies 2u^2 - 4u + v^2 - v = 0. $$
Completing the square, we have
$$ 2u^2 - 4u + v^2 - v = 2(u - 1)^2 - 2 + \left( v - \frac{1}{2} \right)^2 - \frac{1}{4} = 0 $$
which implies that
$$ 2(u-1)^2 + \left( v - \frac{1}{2} \right)^2 = \frac{9}{4}. $$
This is the equation of an ellipse in the $u$-$v$ plane. Manipulating the equality algebraically, we obtain
$$ \left( \frac{\sqrt{2}}{3}(u-1) \right)^2 + \left( \frac{2v - 1}{3} \right)^2 = 1 $$
and so this can be parametrized by letting
$$ \frac{\sqrt{2}}{3}(u - 1) = \cos(\theta), \,\,\, \frac{2v - 1}{3} = \sin(\theta) $$
or, more explicitly,
$$ u = 1 + \frac{3}{\sqrt{2}}\cos(\theta), \,\,\, v = \frac{3 \sin(\theta) + 1}{2} $$
for $\theta \in [0,2\pi]$.
