How do you define "equidistant"?
Here's the definition I assume you are using:
If C is a curve, and E is an ellipse, then C is "equidistant from E" if there is some positive real number r such that for all x in C, inf{|x-y| : y in E} = r.
This is actually an interesting question.
There is a minimum curve which is "inside" all other equidistant curves to E, and I assume this is what you are asking about. This is the curve C such that each point in C has a perpendicular distance of r from the ellipse E. In other words, each point x in C is distance r from some point y on E, and the tangent to y is perpendicular to the line between x and y. I think it's not to hard to show this curve is unique. Moreover, if any point is inside it, it must be closer than r to some point, so C is minimal, in some sense.
Now, let E be parametrized by the equation f(t) = (a*cost, b*sint).
Then f'(t) = (-a*sint, b*cost) -- this is the tangent vector at each point of E.
We are going about the ellipse E counterclockwise, so if we rotate this tangent vector by pi/2 clockwise, we will get a perpendicular vector to the tangent vector that points out of the ellipse.
We can simply use a rotation matrix:
http://en.wikipedia.org/wiki/Rotation_matrix
In this case, theta will be -pi/2 (as we want to rotate clockwise, not counter clockwise). Our matrix will be:
[ cos(-pi/2) -sin(-pi/2)]
[ sin(-pi/2) cos(-pi/2)]
=
[0 1]
[-1 0]
When applied to our tangent vector, we get:
(b*cost,a*sint).
Of course, we need to scale it to be of length r. Let the function s be s(t) = r/sqrt((b*cost)^2 + (a*sint)^2). Note s is real-valued, not vector-valued.
Then our curve C is parametrized by:
g(t) = f(t) + (s(t)*b*cost, s(t)*a*sint)
Fully expanded, we have:
g(t) = [ a*cost + b*cost*r/sqrt((b*cost)^2 + (a*sint)^2), b*sint + a*sint*r/sqrt((b*cost)^2 + (a*sint)^2) ]
So, what you question boils down to is: does g(t) describe an ellipse?
Have fun!
