I think this is a clever solution, however I don't think Excel's solver can handle it. But let me make sure I understand the algorithm.

First, get coordinates $\displaystyle (x_i,y_i),\ i = 1,\ldots,N$ from the picture. Fix a value for $\displaystyle \delta t.$

Next, define a function $\displaystyle Dist(A,\ B,\ a,\ b,\ \delta)$ with the following 4 steps.

- Set $\displaystyle M = 2\pi / (\min(a,b)\delta t).$
- Tabulate $\displaystyle u_k=A\sin(a(k\ \delta t)+\delta),\ v_k=B\cos(b(k\ \delta t))$ for $\displaystyle k=1,\ldots, M.$
- For each $\displaystyle i = 1,\ldots,N$, find a point $\displaystyle (u_j,v_j)$ out of the $\displaystyle M\ (u_k,\ v_k)'s$ which has the minimum distance from $\displaystyle (x_i,y_i)$ to $\displaystyle (u_j,v_j)$.
- Return the sum of the squares of the distances in step 3 as the value of $\displaystyle Dist.$

Finally, minimize $\displaystyle Dist(A,\ B,\ a,\ b,\ \delta).$

I don't think Excel's solver can handle this minimization because I believe it assumes the minimized function is differentiable and numerically approximates the derivatives. But the $\displaystyle Dist$ function, although continuous, will not be differentiable. Another problem with using Excel is that programming the $\displaystyle Dist$ function in a spreadsheet looks difficult.

However with a good programming language and a minimization routine that can handle a non-differentiable function, this could work. Good job!