Well, I am not posting the whole solution.

It'll be better if u try this qun yourself.

I can tell you how to solve this.

Assume co-ordinates A (aCos@,bSin@).

Now, as it's an quilateral triangle,next co-ordinate will be B (aCos(@+120),bSin(@+120))

next will be C(aCos(@+24), bSin(@+240))

Now, you have co-ordinates of equilateral triangle and

I think now you know what to do..

centroid is ((x1+x2+x3)/3,(y1+y2+y3)/3).

Now, it's simple trigo.

still if you have any doubt, you can ask.

Try to do it yourself.

Now, as it's an quilateral triangle,next co-ordinate will be B (aCos(@+120),bSin(@+120))

pl expalin this

Oh!!

that was the wrong solution i posted. I am sorry for that.

here's the correct one:

assume cordinates as A(aCos@,bSin@), B(aCos$,bSin$) and C(aCos&, bSin&).

Now, as you know all the sides of an equilateral triangle are equal.

So, AB=AC, AC=BC and AB=BC.

you'll get the values of all the three variables, since you have three equations.

now, centroid of the triangle is (x1+x2+x3)/3, (y1+y2+y3)/3.

@pranjali: I am considering it to be a general ellipse ((x/a)^2 +(y/b)^2)=1.

Do post again if there's any doubt.

2nd solution:

Assume all the three co-ordinates as mentioned just above.

and then you can use: Tan60 = mode of ((m1-m2)/(1-m1*m2)).

m1,m2 are slopes.

3rd solution:

Assume centroid to be (h,k) and take one point to be (aCos@, bSin@).

now, find the equation of line from these two points.

Now, rotate this line 120 and -120 degree and find the two more intersection points with ellipse.

After that you can just equate the lengths of the side to get the desired result.

I hope it's clear now..

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