+305 An equilateral triangle ABC is inscribed in the ellipse x^2/a^2+y^2/b^2=1, so that B is at (0,b), and line AC is parallel to the x-axis, as shown below. Also, foci F_1 and F_2 lie on sides line BC and line AB, respectively. Determine the value of AB/(F_1*F_2). 
+305 I would appreciate if you gave a solution following your statement. Thanks in advance.
+118687 what do you mean by F_1*F_2
they are points, not distances so it doesn't make sense to me.......
+305 I apologize for the misunderstanding, as I meant the length between the two points F_2 and F_1.
+118687 Since it is a ratio question you can use a convenient ellipse.
Let X be the centre and M be the midpoint of AC
triangle BXF2 and triangle BMA are similar tiranges and they are both 30, 60 90 degree triangles.
So their sides are in the ration 1:2:sqrt3
I want the focal length (c) to be 1 and
I want B to be \((0,\sqrt3)\)
\(c^2=a^2-b^2\\ 1=a^2-3\\ 4=a^2\\ a=2\)
So the ellipse that I will use is
\(\frac{x^2}{4}+\frac{y^2}{3}=1\\ \frac{y^2}{3}=1-\frac{x^2}{4}\)
Equation of line AB
\(y=\sqrt3\;x+\sqrt3\\ y=\sqrt3(x+1)\\ y^2=3(x+1)^2\\ \frac{y^2}{3}=(x+1)^2\\ \)
Solve simultaneously
\(1-\frac{x^2}{4}=(x+1)^2\\ -\frac{x^2}{4}=x^2+2x\\ -x^2=4x^2+8x\\ 5x^2+8x=0\\ x(5x+8)=0\\ x=0\quad or \quad x=-\frac{8}{5} \)
So the x value for A is -8/3
\(AC=AB=BC=\frac{16}{5}\\ F_2F_1=2\\ \frac{AB}{F_1F_2}=\frac{16}{10}=\frac{8}{5} \)
