# Ellipse with non-orthogonal minor and major axes?

If there's an ellipse with non-orthogonal minor and major axes, what do we call it? For example, is the following curve a ellipse?

$x = \cos(\theta)$

$y = \sin(\theta) + \cos(\theta)$

curve $C=\vec(1,0)*\cos(\theta) + \vec(1,1)*\cos(\theta)$

The major and minor axes are $\vec(1,0)$ and $\vec(1,1)$. They are not orthogonal.

Is it still an ellipse?

Suppose I have a point $P(p_1,p_2)$ can I find a point Q on this curve that has shortest euclidean distance from P?

• Is is true that for any point on that curve, the sum of its distances to two fixed points ( the purposed ellipse's foci ) is a fixed number? Can you locate foci at all, to begin with? Aug 23 '12 at 5:31
• You can show that the curve you have is indeed an ellipse, with axes of length $\sqrt{1+\phi}$ and $\sqrt{2-\phi}$, and the major axis inclined at an angle of $\arctan\,\phi$, where $\phi$ is the golden ratio. Aug 23 '12 at 5:45
• The major and minor axes are always orthogonal. This is not obvious, but it is true. Aug 23 '12 at 6:02
• I think what you may not realize is the following: a nonsingular linear transformation (e.g. a shear) always transforms an ellipse to another ellipse, but the axes of the image ellipse are not in general the images of the axes of the original ellipse. Aug 23 '12 at 6:20

More explicitly, we have the decomposition

$$\begin{pmatrix}\cos\,t\\\cos\,t+\sin\,t\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}\cos(t+\eta)\\\sqrt{2-\phi}\sin(t+\eta)\end{pmatrix}$$

where $\tan\,\lambda=\phi$, $\tan\,\eta=1-\phi$, and $\phi=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio. You can check that both your original parametric equations and the new decomposition both satisfy the Cartesian equation $2x^2-2xy+y^2=1$. What the decomposition says is that your curve is an ellipse with axes $\sqrt{1+\phi}$ and $\sqrt{2-\phi}$, with the major axis inclined at an angle $\lambda$.

If we take the linear algebraic viewpoint, as suggested by Robert in the comments, what the decomposition given above amounts to is the singular value decomposition (SVD) of the shearing matrix; i.e.,

$$\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}&\\&\sqrt{2-\phi}\end{pmatrix}\cdot\begin{pmatrix}\cos\,\eta&\sin\,\eta\\-\sin\,\eta&\cos\,\eta\end{pmatrix}^\top$$

The SVD is in fact an excellent way to look at how a matrix transformation geometrically affects points: the two orthogonal matrices on the left and right can be thought of as rotation matrices, reflection matrices, or products thereof, and the diagonal matrix containing the singular values amounts to nothing more than a scaling about the axes of your coordinate system.

Notice that $\sin\theta+\cos\theta \propto \sin(\theta+\pi/4)$, so that the shape you are drawing is just a rotated (and stretched) ellipse (because the "corner" points, $\theta=n\pi/2$, occur with the coordinates shifted from the usual $x$-axis, $y$-axis orientation).

Grapher concurs:

So, in the sense that you've defined it, the axes of the ellipse don't end up non-orthogonal at all -- hopefully that answers the rest of the questions.

In general, a linear combination of sines and cosines can always be written as a single shifted sine or cosine, and thus an ellipse can be rotated (and/or stretched) as such.

Hint: From $y=\sin\theta+\cos\theta$, we get $y-x=\sin\theta$, and therefore $(y-x)^2=\sin^2\theta=1-x^2$. After simplifying and completing the square, can you recognize the curve?

The major and minor axes do turn out to be orthogonal.