# The equation for a rotated ellipse

So, I recently found an interesting way to find the equation of a rotated ellipse. As we know, an ellipse is characterized by the equation

Where a is the length of the semimajor axis and b is the length of the semi minor axis.

A rotated ellipse is exactly what it says. It looks like this

Today, we’ll try to derive the formula for an arbitrary rotated ellipse, that is an ellipse with semimajor and minor axes of lengths a and b rotated by an angle θ. What’s unique about this approach is that firstly, it looks at the ellipse from a 3-D point of view rather than 2-D, and secondly, it uses concepts from simple harmonic motion.

Let’s take an arbitrary ellipse

Look at the equation highlighted in yellow. This can be interpreted as the intersection of 2 3-D graphs

Lets look at this visually. An ellipse can be thought of as an intersection of a plane and a paraboloid

Now we do something unusual. We assume that the function z(x,y) describes the potential energy of an object. The force field associated with this potential is

The eigenvalues of this matrix are -2b² and -2a² respectively. The respective eigenvectors lie on the lines y=0 and x=0 respectively.

Now’s the important part. The force field above implies that if a particle is placed on one of the eigenvectors, then it undergoes simple harmonic motion. Therefore, simply rotating the potential energy graph will not change the nature of the motion, only the direction. So even if the graph of z(x,y), and consequently the force field, are rotated by an angle θ, the eigenvalues will remain the same, while eigenvectors change. In fact, we can find out how the eigenvectors change.

The rotated eigenvectors will be perpendicular, and it’s important to realize that the eigenvectors will be along the semimajor and semiminor axes. Thus, the lines containing the eigenvectors will becomes y=tanθx and y=-cotθx. The eigenvalues will remain the same (-2b2 and -2a2 respectively). The rotated z(x,y), which I’ll call z’(x,y), will represent a new force field F’.

Why have I assumed the components of the force field to be linear? Because the rotated graph is still a paraboloid, with quadratic terms. Differentiating obviously gives linear terms.

The matrix in (1) will have the same eigenvalues mentioned before, but with different eigenvectors. Thus,

This will give us 4 equations in α, β, γ and δ.

The solutions of these equations are:

As expected, β=γ, as the field must be conservative. Then, we can use the facts

To derive z’. Thus,

This is the function z’(x,y), which represents the rotated paraboloid curve.

Now its important to realize that the graph above (z’(x,y)) is simply the original z(x,y) rotated. This means that we simply have to equate this function to z=a2b2 to find the equation for our rotated ellipse.

So, the equation of an ellipse rotated by an angle θ becomes

So that’s it! That’s the equation for a rotated ellipse! The derivation is a little convoluted, but in the end, it leads to a rather nice result! Also, I thought it was interesting to combine multivariable calculus with a little bit of physics. Bye!

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