Two ellipses with constant gap

Question

I want to have two ellipses with constant radial gap in between. Basically if the ellipse 1 is:

$$r_1\left( \theta\right)=\frac{1}{\frac{{\cos^2\theta}}{a_1^2}+\frac{{\sin^2\theta}}{b_1^2}}$$

and ellipse 2 is

$$r_2\left( \theta\right)=\frac{1}{\frac{{\cos^2\theta}}{a_2^2}+\frac{{\sin^2\theta}}{b_1^2}}$$

Now for each $a_1$ and $b_1$ how can I find $a_2$ and $b_2$ to have

$$r_2\left( \theta\right)=r_1\left( \theta\right)+c$$

P.S. Even approximate solutions would also do. I mean even if there is no analytical solution for a constant $c$ radial gap between two ellipses, maybe there is a $c(\theta)$ where it's variation is minimum from 0 to $2\pi$. In other words if we define average gap size as:

$$c_{avg}=\frac{1}{2\pi}\int^{2\pi}_0\left( r_2\left( \theta\right)-r_1\left( \theta\right) \right)d\theta$$

How to find the $a_2$ and $b_2$ in a way to minimize absolute deviation of gap size from average:

$${d}_{abs}=\frac{1}{2\pi}\int^{2\pi}_0\left| c\left( \theta\right)-c_{avg} \right|d\theta$$

Answer 1

There is an analytic expression for a constant radial gap around an ellipse. You start by writing the ellipse equation and getting its gradient. So for example, let the ellipse be $$\frac{x^2}{1}+\frac{y^2}{2}=1.$$ Then its gradient will be $$\nabla=\left(\begin{array}{c}2x\\y\end{array}\right)$$ We will use that in a bit, but understand that it gives the normal direction away from the ellipse for every point. Next convert the ellipse equation to parametric form. $$\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}CP_{x}\\CP_{y}\end{array}\right)+\left(\begin{array}{c}a\cdot cos(t)\\b\cdot sin(t)\end{array}\right)$$ In this case, I let $a=1$ and $b=\sqrt{2}$ so we get $$\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}0\\0\end{array}\right)+\left(\begin{array}{c}cos(t)\\\sqrt{2}sin(t)\end{array}\right)$$ Next, I just use the gradient to write the new locus of points that are a constant distance, $r$, from the existing ellipse.

$$\left(\begin{array}{c}x_{1}\\y_{1}\end{array}\right)=\left(\begin{array}{c}0\\0\end{array}\right)+\left(\begin{array}{c}cos(t)\\\sqrt{2}sin(t)\end{array}\right)+r\frac{\left(\begin{array}{c}2\cdot cos(t)\\\sqrt{2}sin(t)\end{array}\right)}{\left\Vert \left(\begin{array}{c}2\cdot cos(t)\\\sqrt{2}sin(t)\end{array}\right)\right\Vert }$$ Hopefully, my picture shows an example for this ellipse with $r=1/4$. I am pretty sure that the new locus is not an ellipse.