How to parametrize the ellipse $4x^2+\frac{y^2}9+xy=36$?
By Matthew Cannon
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?
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$\begingroup$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:$$ {3\sqrt7\over2} x = 18\cos t \implies x={12\over\sqrt7}\cos t\\ {9\over2}x+y = 18\sin t \implies y=18\sin t-{54\over\sqrt7}\cos t. $$
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