Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $c = 0$, the parametric form is obviously $x = \sqrt{\frac{d}{a}} \cos(t), y = \sqrt{\frac{d}{b}} \sin(t)$.

When $c \neq 0$ the sine and cosine should be phase shifted from each other. How do I find the angular shift and from there how do I adjust the factors multiplying the sine and cosine?

share|cite|improve this question
Principal axis theorem gives you an answer in the sense that there is a worked out example there. Hopefully somebody has the time to give you a bit more theory. I gotta go :-( – Jyrki Lahtonen Oct 5 '11 at 14:03
Formulae 16-23 here are useful. – J. M. Oct 5 '11 at 16:05
up vote 8 down vote accepted

You would complete squares: $\left(a x + \frac{1}{2} c y\right)^2 + \left(a b - \frac{c^2}{4} \right) y^2 = a d$.

From there: $a x + \frac{c}{2} y = \sqrt{a d} \sin(t)$ and $\sqrt{a b - \frac{c^2}{4}} y = \sqrt{a d} \cos(t)$, assuming $c^2 < 4 a b$, and $a d > 0$.

Solving for $x$ and $y$ and denoting $\mathcal{D} = 4 a b - c^2$ $$ x(t) = \sqrt{\frac{d}{a}} \left( \sin(t) - \frac{c}{\sqrt{\mathcal{D}} } \cos(t) \right) \qquad y(t) = \frac{2 \sqrt{a d}}{\sqrt{\mathcal{D}}} \cos(t) $$

share|cite|improve this answer
I like that. +2 if I could. Much appreciated. – jnm2 Oct 5 '11 at 15:33
+1 You didn't use the principal axes, but found a parametrization anyway :-) – Jyrki Lahtonen Oct 5 '11 at 15:48
In the Olden Days (when I was in school) there was a college course called "Analytic Geometry". It would include such things as finding the rotation to eliminate the $xy$ term in a plane conic section. – GEdgar Oct 5 '11 at 18:55

If the parametric ellipse coordinates are $\left(x(t),y(t)\right) = (X \cos\varphi \cos(t)-Y \sin\varphi \sin(t), Y \cos\varphi \sin(t)+X \sin\varphi \cos(t)) $

Then the parameters $ X,Y,\varphi $ are $$ X =\pm \sqrt{\frac{2 d}{a+b+\sqrt{(a-b)^2+c^2}}} $$

$$ Y =\pm \sqrt{\frac{2 d}{a+b-\sqrt{(a-b)^2+c^2}}} $$

$$ \varphi = \frac{1}{2}\tan^{-1}\left(\frac{c}{a-b}\right) $$


$10 x^2+20 y^2-18 x y = 100$

The coefficients are ($a=10$, $b=20$, $c=-18$, $d=100$)

Wolfram Alpha Implicit Curve

The parametric coefficients are $(-1.008311 \cos(t)+3.973667 \sin(t), 1.713639 \cos(t)+2.338119 \sin(t))$ Wolfram Alpha Parametric Curve

share|cite|improve this answer

(This is supposed to be a comment on ja72's answer, but it got too long.)

ja72's answer mentions the formula

$$\tan\,2\varphi = \frac{c}{a-b}$$

I consider it a bit wasteful of effort to evaluate an arctangent and the subsequently pass it as the argument of a trigonometric unction much later; to avoid this, we can use the double angle formula for the tangent and the "Citardauq" formula in tandem to yield the relation


Denoting this expression as $t$, we can then substitute this into the rotation matrix


ja72's answer already gave the formulae for the axes; remember that $(p\sin\,u,q\cos\,u)$ and $(p\cos\,u,q\sin\,u)$ are the same ellipse traversed differently, so you can subsequently rotate whichever of the two expressions you pick with the rotation matrix I gave earlier.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.