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Arbitrary ellipses means that they can be scaled, translated and rotated in any way in 2D. Do you know some high-school method (might be slightly more advanced than that) to find the minimum distance? I'd love a symbolic expression/solution but a numerical solution for a specific pair of ellipses would be greatly appreciated too.Thanks very much for any solution.

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possible duplicate of minimum distance between two ellipses – joriki Sep 12 '12 at 13:38
up vote 3 down vote accepted

In general, an ellipse is given by $$G(x,y)=a x^2 + b y^2 + c x y + d x + e y + f =0.$$

Let us denote the two ellipses with the subscripts 1 and 2. As we would like to minimize the square distance $d^2=(x_1 - x_2)^2 + (y_1 + y_2)^2$ between the two ellipses. We use the method of Lagrange multiplier and write $G = d^2 + \lambda_1 G_1(x_1,y_1) + \lambda_2 G_2(x_2,y_2)$ with the conditions for an extremum $$ \partial_{x_1} G = 2 (x_1 - x_2) + λ_1(2 a_1 x_1+ c_1 y_1 +d_1) =0,$$ $$\partial_{y_1} G =2 (y_1 - y_2) + λ_2(2 b_1 y_1+ c_1 x_1 +e_1) =0,$$ $$ \partial_{x_2} G = 2 (x_2 - x_1) + λ_1(2 a_2 x_2+ c_2 y_2 +d_2) =0,$$ and $$\partial_{y_2} G =2 (y_2 - y_1) + λ_2(2 b_2 y_2+ c_2 x_2 +e_2) =0.$$ Together with the conditions $G_1 = G_2 =0$, we have 6 equations for 6 unknowns (though some are not linear but quadratic). We can "easily" solve for $x_1,y_1,x_2,y_2,\lambda_1,\lambda_2$. Plugging the different solutions into the expression for $d^2$, we can select the one that minimizes the distance.

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Do you think you could show me how to solve for x1,y1,x2,y2,λ1,λ2? Because essentially I got up to this stage every time and got stuck and couldn't proceed. It might be easy as you say but I just can't come to the solution. – David Hoffman Sep 10 '12 at 19:29
Laplace multiplier? Did you mean Lagrange multiplier? – Rod Carvalho Sep 10 '12 at 19:32
Yes, that's certainly what he meant. He just made a little mistake. – David Hoffman Sep 10 '12 at 19:34
So basically, what I have to do now is to express one variable in the terms of the others, and gradually decrease the number of variables until i get the solution for one, and then plug it in for the others, right? – David Hoffman Sep 10 '12 at 19:49
In general, I think you'd end up solving a polynomial of degree $12$, so numerical methods will be needed. Certainly it's beyond what's usually considered as "high school". – Robert Israel Sep 10 '12 at 20:20

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