Locus of centroid of equilateral triangle inscribed in ellipse.

Question

Problem

Find the locus of the centroid of an equilateral triangle inscribed in the ellipse $x^2 / a^2 + y^2 / b^2 = 1$

My attempt

I assumed 3 parametric points on ellipse P, Q and R. And assumed the centroid to be $G(h,k)$ As we know the formula for centroid (taking average of three coordinate assumed). Hence got the relation $$ h= a(cos \theta_{1} + cos \theta_{2} + cos \theta_{3} )/3 $$ $$ k = b(sin \theta_{1} + sin \theta_{2} + sin \theta_{3} )/3 $$ Now I tried using formula for circumcentre and no matter whicher formula I used I used to get the same above equation. I tried writing equation of three sides and manually solving the centroid.

I also got an intuition that in an equilateral triangle all 4 centres are same. So centroid and circumcentre are same. So $ PG = QG = RG $. So I wrote down all the equation and equated them by comparing the LHS equation and RHS equation to get $$ cos \theta_{1} = cos \theta_{2} = cos \theta_{3} $$ $$ sin \theta_{1} = sin \theta_{2} = sin \theta_{3} $$ which is obviously wrong as that would mean all the tree points assumed intersected into one single point.

Where exactly am I wrong? What did I miss that I had to think about which I didn't think. I am looking for some nice elegant human solution to this problem as I have seen a similar question in mathematical stack exchange in which they have directly solved equations and got very big equation for h and k.

Below I attach solution that is provided in textbook but in the textbook solution there is a formula written directly. If anyone could explain how I could prove the formula as no information is provided from where the formula came.Solution 1. Solution 2

Thanks for helping. I have been working on this for 3 days with no improvement. This is my first question on mathematics stack exchange so any comments on how should I state something is appreciated if I stated the wrong way.

Answer 1

If you start with the coordinates of the vertices of the triangle as: $$ A=(a\cos\alpha,b\sin\alpha),\quad B=(a\cos\beta,b\sin\beta),\quad C=(a\cos\gamma,b\sin\gamma), $$ then you can find the coordinates of the circumcenter as the intersection of the perpendicular bisectors of two sides. The equation of the perpendicular bisector of side $AB$ is: $$ y-\frac{b}{2} (\sin \alpha+b \sin\beta) =\frac{a \cos \alpha-a \cos\beta}{b\sin\beta-b \sin \alpha} \left(x-\frac{a}{2} (\cos \alpha+\cos \beta)\right) $$ and you can find with the obvious substitutions the equation of the perpendicular bisector of side $BC$. Solving the system of those two equations gives a quite long solution, which can be put into the simple form: $$ x= \frac{a^2-b^2}{a}{ \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha+\gamma}{2} \cos \frac{\beta+\gamma}{2}}, \quad y= \frac{b^2-a^2}{b}{ \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha+\gamma}{2} \sin \frac{\beta+\gamma}{2}}. $$ Finally, using trig addition formulas, these can be rewritten in the form given by your textbook: $$ \begin{align} x&= \frac{a^2-b^2}{4a} [\cos \alpha+\cos\beta+\cos\gamma+\cos(\alpha+\beta+\gamma)], \\ \ \\ y&= \frac{b^2-a^2}{4b} [\sin \alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)]. \end{align} $$

EDIT. In your textbook this formula is derived in a smarter way, from the property that the eccentric angles of the four intersection points between an ellipse and a circle add up to a multiple of $2\pi$. See pages 497-498.