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?

https://www.sparknotes.com/math/precalc/conicsections/section5/#:~:text=To%20eliminate%20the%20xy%20term,that%20cot(2%CE%B8)%20%3D%20 — saulspatz, Oct 07 '20 at 16:20
$$\sqrt{(x+\frac{\sqrt{2}}{\sqrt{3}})^2+(y-\frac{\sqrt{2}}{\sqrt{3}})^2-}+\sqrt{(x-\frac{\sqrt{2}}{\sqrt{3}})^2+(y+\frac{\sqrt{2}}{\sqrt{3}})^2}=2\sqrt{2}$$ $$\frac32((x+\frac{\sqrt{2}}{\sqrt{3}})^2+(y-\frac{\sqrt{2}}{\sqrt{3}})^2-\frac23(y-x-\sqrt{6})^2/2)=0$$
$$\frac32((x-\frac{\sqrt{2}}{\sqrt{3}})^2+(y+\frac{\sqrt{2}}{\sqrt{3}})^2-\frac23(y-x+\sqrt{6})^2/2)=0$$ — Jan-Magnus Økland, Oct 07 '20 at 17:51

José Carlos Santos · Accepted Answer · 2020-10-07T16:40:03.920

3

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?

edited Oct 07 '20 at 16:40

answered Oct 07 '20 at 16:32

José Carlos Santos

427,504

I see now that this is a solution. But why did you immediately put $\frac {1}{\sqrt 2 }$? – Adriano Banchieri Oct 07 '20 at 16:57
1

I drew the curve and I saw that the axes of this ellipse made a $45^\circ$ angle with respect to the usual axes. So, I did a $45^\circ$ degrees rotation. Note that both the sine and the cosine of that angle are equal to $1/\sqrt2$. – José Carlos Santos Oct 07 '20 at 17:13

score 1 · Answer 2 · answered Oct 07 '20 at 16:41

$$x^2+y^2-xy=1 \implies \frac{1}{2}[(x+y)^2+(x-y)^2]+\frac{1}{4}[(x+y)^2-(x-y)^2]=1$$ $$\frac{3(x+y)^2}{4}+\frac{(x-y)^2}{4}=1$$ $$\implies \frac{3[(x+y)/\sqrt{2}]^2}{2}+\frac{[(x-y)/\sqrt{2}]^2}{2}=1 \implies \frac{X^2}{2/3}+\frac{Y^2}{2}=1$$ 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 $\sqrt{2/3}$.

score 1 · Answer 3 · answered Oct 07 '20 at 20:03

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.

score 1 · Answer 4 · answered Oct 07 '20 at 20:36

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 $a=\sqrt{2/3}$.

Rotated ellipse: how to work with it

4 Answers4