how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$

Question

I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it. Starting from the matrix:

$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3 \end{array}\right]$$

I computed the eigenvalues $2 \pm \frac{1}{2}\sqrt{5}$ and the eigenvectors (not normalized):

$$\left[ \begin{array}{c} x\\ y \end{array}\right] = \left[ \begin{array}{c} 1\\ 2\pm \sqrt{5} \end{array}\right] $$

So then I tried writing down some combination of the data I generated:

$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right] = \cos \theta \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right] + \sin \theta \left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right] $$

However, I have a hard time checking the ellipse equation holds true for all $\theta$:

$$x(\theta)^2 + x(\theta)y(\theta) + 3y(\theta)^2 = 1$$

What are the correct functions $x(\theta), y(\theta)$ ?

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Following the comments, rescaling the eigenvectors and multiplying the eigenvalues:

$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right] = \frac{2 + \frac{1}{2}\sqrt{5}}{\sqrt{10 + 4 \sqrt{5}}} \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right]\cos \theta + \frac{2 - \frac{1}{2}\sqrt{5}}{\sqrt{10 - 4 \sqrt{5}}}\left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]\sin \theta $$

Is it clear that the ellipse equation is satisfied? I am not sure how to check this.

Answer 1

You have $x^2 +xy + 3y^2=(x+1/2 y)^2+11/4 y^2=1$.

You can then take: $$\begin{array}{lll} \sin \theta & = & x+1/2 y\\ \cos \theta &= &\sqrt{11}/2 y \end{array}$$

Which is equivalent to: $$\begin{array}{lll} y &= 2/\sqrt{11} \cos \theta\\ x &=\sin \theta - 1/\sqrt{11} \cos \theta \end{array}$$

to get a parametrization like the one you're looking for.

Answer 2

If you need a parametrization, it is best to just complete the square, rather then exploiting the full power of the spectral theorem. $$ x^2+xy+3y^2 = 1 $$ is equivalent to: $$ \left(2x+y\right)^2 + 11 y^2 = 4$$ hence $2x+y=2\cos\theta,y=\frac{2}{\sqrt{11}}\sin\theta$ is a valid parametrization, that leads to: $$ x = \cos\theta-\frac{1}{\sqrt{11}}\,\sin\theta,\quad y=\frac{2}{\sqrt{11}}\,\sin\theta.$$

Answer 3

let us define $\theta$ by $$\cos\theta = \frac1{\sqrt{10+4\sqrt 5}}, \sin \theta=\frac{2+\sqrt 5}{\sqrt{10+4\sqrt 5}}.$$ then you can verify that $$\pmatrix{1&1/2\\1/2&3}\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta} = \pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}\pmatrix{2+\sqrt5/2&0\\0&2-\sqrt5/2}$$ that is $$\pmatrix{1&1/2\\1/2&3}= \pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta} \pmatrix{2+\sqrt5/2&0\\0&2-\sqrt5/2} \pmatrix{\cos \theta&\sin \theta\\-\sin \theta&\cos \theta}$$ that is $$A = U^\top DU, U^\top U = I \text{ where } U = \pmatrix{\cos \theta&\sin \theta\\-\sin \theta&\cos \theta} . $$

you can now define the new coordinate transformation by $\xi, \eta$ by the relation $$\pmatrix{\xi\\\eta} = U\pmatrix{x\\y}, \pmatrix{x\\y} = U^\top \pmatrix{\xi\\\eta}.$$ with this we get $$\begin{align}x^2 + xy + 3y^2 &=\pmatrix{x&y} A\pmatrix{x\\y}\\ &= \pmatrix{x&y} U^\top D U \pmatrix{x\\y}\\ &= \pmatrix{\xi & \eta}D\pmatrix{\xi\\\eta} \\ &= (2 + \sqrt5/2)\xi^2 + (2 - \sqrt5/2)\eta^2\end{align}$$