How to parametrize a shifted and tilted ellipse from its quadratic equation

Question

I have the ellipse given by $-3x^2-3y^2+4xy=-2$.

First question is, how do I know that it is really an ellipse? Wolfram alpha says so, but how could I tell in advance?

But more importantly, how can I find a suitable parametrization for this ellipse? I'm aware o similar questions in this forum, where rather convoluted expressions are given for the parameters of a rotated ellipse. But I don't see how to match this to my problem of the search of the parametrization.

For example, in non-tilted ellipses you have to manipulate the equation trying to get the canonical form $\left(\frac{x-x_0}{a}\right)^2 + \left(\frac{y-y_0}{a}\right)^2=1$. Once you have $a, b, x_0$ and $y_0$, you just apply:

$x=x_0+a\cos \theta$

$y=y_0+b\sin\theta$

with $0\le\theta\le2\pi$ and you have it.

Now, how can I do something similar to find the $x(\theta), y(\theta)$ for a tilted ellipse?

Answer 1

The equation is symmetric in $x$ and $y$, so it is natural to try the change of variable

$$x:=u+v,\\y:=u-v.$$ (This is a rotation by $45°$ combined with a scaling by $\sqrt2$, i.e. a similarity transform.)

Plugging in the equation, we get

$$-6(u^2+v^2)+4(u^2-v^2)=-2$$ or

$$u^2+5v^2=1.$$

This is an ellipse centered at the origin, with axes of length $1$ and $\dfrac1{\sqrt5}$ (after scaling).

Answer 2

Recognize that you could use the trick

$$(u+v)(u-v)=u^2-v^2$$

to decouple a pair of coupled variables, such as $xy$. So, let

$$x=u+v, \>\>\> y = u-v$$

to recast

$$-3x^2-3y^2+4xy=-2$$

into the canonical ellipse form

$$u^2+5v^2=1$$

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Edit:

This maneuvering would also work in the general case where $x^2$ and $y^2$ have different coefficients. For example, if the equation is instead,

$$-3x^2-2y^2+4xy=-2$$

Then, let

$$x = u + v, \>\>\> y=u-5v$$

Its canonical form becomes,

$$u^2+73v^2=2$$

Answer 3

As mentioned in other answers, this case is relatively simple because the symmetry of the equation leads immediately to the principal axes being parallel to the vectors $(1,1)$ and $(-1,1)$, which then gets you a parameterization that uses these principal axes of the ellipse. More generally, you can work out the required rotation directly.

First thing’s first, however: how do you tell that the equation represents an ellipse? Examine its discriminant, which in this case is $4^2-4(-3)(-3)=-32\lt0$, which indicates an ellipse. If it were positive, you’d have a hyperbola, while if zero a parabola. (I’m ignoring the possibility of a degenerate conic here.)

Now, we want to find an angle of rotation $\phi$ that eliminates the cross term in $xy$. Applying the rotation to the general conic equation $Ax^2+Bxy+Cy^2+\dots=0$ (the linear and constant terms don’t contribute to this) leads to the condition $$2(A-C)\cos\phi\sin\phi+B(\cos^2\phi-\sin^2\phi)=0,$$ or $$\tan{2\phi}=\frac B{C-A}.$$ For your ellipse, $C-A=0$, therefore $\phi=\pm\pi/4+n\pi.$ Note that you don’t really need to know $\phi$ itself to apply the rotation, so using half-angle formulas you can get explicit expressions for $\sin\phi$ and $\cos\phi$ in terms of $A$, $B$ and $C$.

Alternatively, for a parameterization of the form $\mathbf u\cos t+\mathbf v\sin t+\mathbf c$, the vectors $\mathbf u$ and $\mathbf v$ don’t need to be the principal half-axes of the ellipse: any pair of conjugate half-diameters will do. These can be found by choosing any convenient value for $\mathbf u$ and then using the fact that the conjugate diameter is parallel to the tangent at the point $\mathbf u+\mathbf c$.

Finally, there are other ways to parameterize an ellipse. You might find a rational parameterization more convenient for some purposes. For instance, let $p_0$ be a point on the ellipse that has an extreme $x$-value so that the tangent at that point is vertical. Compute the other intersection of the line through $p_0$ has slope $m$ with the ellipse. This gives you a parameterization in terms of the slope of the chords through $p_0$.