How to parametrize the ellipse $4x^2+\frac{y^2}9+xy=36$?

Question

I know how to parametrize an ellipse that looks like $4x^2+\dfrac{y^2}9=36$ by using polar coordinates:$$2x=6\cos t\Longrightarrow x=3\cos t\\\frac y3=6\sin t\Longrightarrow y=18\sin t$$ But can someone explain how would I go about parametrizing the following ellipse? $$4x^2 + \frac{y^2}9 + xy = 36$$ Can I use polar coordinates?

Answer 1

You can "complete the square". First of all rewrite your equation without denominators: $$ 36x^2+9xy+y^2=18^2, $$ then observe that $$ 36x^2+9xy+y^2=\left({9\over2}x+y\right)^2+\left(36-{81\over4}\right)x^2= \left({9\over2}x+y\right)^2+\left({3\sqrt7\over2} x\right)^2. $$ Hence you can rewrite your equation as: $$ \left({9\over2}x+y\right)^2+\left({3\sqrt7\over2} x\right)^2=18^2 $$ and a possible parametrisation is thus: $$ {3\sqrt7\over2} x = 18\cos t \implies x={12\over\sqrt7}\cos t\\ {9\over2}x+y = 18\sin t \implies y=18\sin t-{54\over\sqrt7}\cos t. $$