Rotated ellipse: how to work with it

Question

I read this post about rotated ellipses and their equation Rotated Ellipse, but I'm still puzzled. I came across $$ x^2 +y^2 +xy - 1=0$$ At first I wasn't thinking of it as an ellipse, but my book said so. I tried to calculate the usual parameters of the ellipse (a, b), its centre, and the angle of rotation with the post I quoted before but it didn't help. What do you suggest in such cases? How can I figure out a canonical equation?

Answer 1

Note that, if $x=\frac1{\sqrt2}(X+Y)$ and $y=\frac1{\sqrt2}(X-Y)$, then\begin{align}x^2+y^2+xy-1=0&\iff\left(\frac{X+Y}{\sqrt2}\right)^2+\left(\frac{X-Y}{\sqrt2}\right)^2-\frac{(X+Y)(X-Y)}2=1\\&\iff\frac32X^2+\frac12Y^2=1.\end{align}Can you take it from here?

Answer 2

$$x^2+y^2-xy=1 \implies \frac{1}{2}[(x+y)^2+(x-y)^2]+\frac{1}{4}[(x+y)^2-(x-y)^2]=1$$ $$\frac{3(x+y)^2}{4}+\frac{(x-y)^2}{4}=1$$ $$\implies \frac{3[(x+y)/\sqrt{2}]^2}{2}+\frac{[(x-y)/\sqrt{2}]^2}{2}=1 \implies \frac{X^2}{2/3}+\frac{Y^2}{2}=1$$ This is the orthogonalized form of ellipse with axes as $X=0$ as major and $Y=0$ as minor axis.Length of semi-major axis is $\sqrt{2}$ and minor axis is $\sqrt{2/3}$.

Answer 3

When you have your generalized quadratic.

$Ax^2+Bxy + Cy^2 + \cdots = 0$

$\frac {C-A}{B} = \cot 2\theta$ will give you the angle of rotation.

In this case:

$0 = \cot 2\theta$

And $\theta = \frac {\pi}{2}$

Then you make the substitution:

$x = x'\cos\theta + y'\sin\theta\\ y = -x'\sin\theta + y'\cos\theta$

And simplify.

Answer 4

By inspection one finds that points $$ (\pm1,0);\quad (0,\pm1);\quad (\pm 1/\sqrt3,\pm 1/\sqrt3); $$ all belong to the conic. This is clearly an ellipse centred at $(0,0)$ and with line $y=-x$ as major axis. Its foci must then lie at $(\pm a,\mp a)$ for some $a$. Imposing that the sum of the distances of the foci from $(\pm1,0)$ is the same as the sum of the distances of the foci from $(1/\sqrt3,1/\sqrt3)$ one finds an equation for $a$, with solution $a=\sqrt{2/3}$.