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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?

Ѕᴀᴀᴅ
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Abdirahman Mohamed
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    The idea would be to convert that ellipse into a "standard form" one, that is, one that looks something like $a^2(Ax + By)^2 + b^2(Bx - Ay)^2 = 1$ with $A^2 + B^2 = 1$. After that, you solve for the centre $(x_0, y_0)$ and parameterise as earlier. – Aryaman Maithani May 12 '20 at 13:55
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    $\newcommand{\x}{\mathbf{x}}$Let $f(x,y)=4x^2+xy + \frac{y^2}{9}$, then $f(x,y) = \x^T A \x$ where $\x = \begin{bmatrix}x \ y\end{bmatrix}$ and $A = \begin{bmatrix}4 & 1/2 \ 1/2 & 1/9\end{bmatrix}$. You can then diagonalise $A$ and go from there. You can find an example at https://math.stackexchange.com/questions/280937/finding-the-angle-of-rotation-of-an-ellipse-from-its-general-equation-and-the-ot. – Minus One-Twelfth May 12 '20 at 14:32
  • Be warned that parameter $t$ in your equations is NOT the angle of polar coordinates. – Intelligenti pauca May 12 '20 at 15:12
  • What sort of parameterization would you like? There are many possibilities. – amd May 12 '20 at 19:30
  • Beware, the standard form route leads to the parametrization $$x= \frac{81 ,2^{\frac{3}{2}},\sin t}{\sqrt{2612-70,\sqrt{1306}},\sqrt{ \sqrt{1306}+37}}+\frac{81,2^{\frac{3}{2}},\cos t}{\sqrt{37- \sqrt{1306}},\sqrt{70,\sqrt{1306}+2612}}$$

    $$y=\frac{9,2^{\frac{ 3}{2}},\left(\sqrt{1306}-35\right),\sin t}{\sqrt{2612-70, \sqrt{1306}},\sqrt{\sqrt{1306}+37}}-\frac{9,2^{\frac{3}{2}}, \left(\sqrt{1306}+35\right),\cos t}{\sqrt{37-\sqrt{1306}}, \sqrt{70,\sqrt{1306}+2612}} $$

     – Jan-Magnus Økland May 14 '20 at 08:27

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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: $$ {3\sqrt7\over2} x = 18\cos t \implies x={12\over\sqrt7}\cos t\\ {9\over2}x+y = 18\sin t \implies y=18\sin t-{54\over\sqrt7}\cos t. $$

Intelligenti pauca
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