Finding point on ellipse equally distant from two other points on the ellipse

Question

I have an ellipse with two points on it: A and C (with known coordinates). Point O is the center of the ellipse (coordinates are given).

I need to find coordinates of point B which also lies on the ellipse, and it must be equally distant from points A and C (lengths of line segments AB and BC must be equal).

I have tried to solve this myself, but without success. Could you please suggest any solution?

Thank you in advance! UPDATE: the parametric equation of the ellipse is alse given.

Answer 1

Since you want equal lengths for the line segments, rather than for the two arc lengths, this is solvable, if a bit messy. Suppose the ellipse parametric equation is $x=a\cos t,\ y=b \sin t.$ Further suppose the two given points on the ellipse are $A=(x_1,y_1)$ and $C=(x_2,y_2).$ [Note after this will use $C,S$ for the desired values $\cos t, \sin t$, so that the coordinates of $B$ are $(aC,bS).$ The difference of squared segment lengths is then set to zero: $$(aC-x_1)^2+(bS-y_1)^2-(aC-x_2)^2-(bS-y_2)^2=0.$$ Now the quadratic terms in the unknown $C,S$ cancel and we arrive at a linear relation between $C,S:$ $$[2a(x_2-x_1)]C + [2b(y_2-y_1)]S=x_2^2-x_1^2+y_2^2-y_1^2.\tag{1}$$ We have another equation $C^2+S^2=1$ to use, and with $(1)$ this lets us solve for $C,S.$ There will be solutions for this, given we have the original points $A,C$ on the ellipse, for geometric reasons. It's a bit involved to solve $(1)$ and $C^2+S^2=1$ if we leave all the letters in it. However for any particular values a "closed form" answer for the sine and cosine of $t$ can be found. And once that is known we can get coordinates of the unknown point $B.$

Answer 2

The parametrization will most probably be of the form $$t\mapsto {\bf z}(t)=\cos t\ {\bf p}+\sin t\ {\bf q}\qquad(0\leq t\leq 2\pi)\tag{1}$$ for given vectors ${\bf p}$ and ${\bf q}$. A point ${\bf z}$ that is equidistant from ${\bf a}$ and ${\bf c}$ lies on the perpendicular bisector of these two points. It satisfies the equation $$2{\bf z}\cdot({\bf c}-{\bf a})=|{\bf c}-{\bf a}|^2\ .$$ Now plug this into $(1)$ and obtain an equation for $t$ that will have two solutions.