Modeling path of a rolling ellipse I'm trying to solve Project Euler problem 525. My approach is to find a parametric equation that can model the path of the center point as it rolls, then take the arc length of that function for one rotation.

A rotated ellipse can be expressed with the equation: $$\frac{(x\cos\theta-y\sin\theta)^2}{a^2}+\frac{(x\sin\theta+y\cos\theta)^2}{b^2}=1$$

And in order to make this ellipse lie tangent or "rest" on the x axis when rotated $\theta$ degrees, there is a y shift of: $\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$. Here is a visual representation of this. Therefore the y component of my final equation will be: $y(\theta) = \sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$.

The x component of the equation is harder. After $\theta$ degrees of rotation, the point of the ellipse touching the x axis will have traveled a distance equal to the arc length of the ellipse from 0 to $\theta$. This can be expressed with the arc length formula using the parametric form of an ellipse:

$$L(\theta) = \int_{0}^{\theta} \sqrt{y'(\theta)^2 + x'(\theta)^2} d\theta$$

This almost solves it, but there is an additional horizontal distance between the tangent point and the center point as seen in the picture below. 



How do I express this extra x-distance? I think it might not even require any calculus.

(Note, I don't want a solution to the problem, that takes the fun out of the problem. Just some help to get through this step or a point in the right direction.)
One year after you, i'm triyng to solve this problem. I hope you have not yet found the solution...
Please see here a picture: https://imgur.com/a/pcvfBrz

Coordinates of M (acos(t);b(1+sin(t))
Tangent vector T(-a.sin(t);b.cos(t))
Normal vector N(b.cos(t), a.sin(t))

You can then search the coordinates of I = intersection of (MI) and (CI)

With w = (a²sin²(t)+b²cos²(t)) then 
I = ( (a-ab/w).cos(t) , (b-ab/w).sin(t) + b )

after you can easily calculate MI and CI which are the values ​​you are looking for.

But after, when you try to calculate the length , it's more complicated.

(dMI/dt)²=a²b²(b²-a²)²sin²(t)cos²(t)\w^3

(dOM/dt+dIA/dt)²=(sqr(w)+(w²-a²b²)/w^(3/2))²

and then... i'm sucked...
One year after you, i'm triyng to solve this problem. I hope you have not yet found the solution...
Please see here a picture: https://imgur.com/a/pcvfBrz

Coordinates of M (acos(t);b(1+sin(t))
Tangent vector T(-a.sin(t);b.cos(t))
Normal vector N(b.cos(t), a.sin(t))

You can then search the coordinates of I = intersection of (MI) and (CI)

With w = (a²sin²(t)+b²cos²(t)) then 
I = ( (a-ab/w).cos(t) , (b-ab/w).sin(t) + b )

after you can easily calculate MI and CI which are the values ​​you are looking for.

But after, when you try to calculate the length , it's more complicated.

(dMI/dt)²=a²b²(b²-a²)²sin²(t)cos²(t)\w^3

(dOM/dt+dIA/dt)²=(sqr(w)+(w²-a²b²)/w^(3/2))²

and then... i'm sucked...