An equilateral triangle $PQR$ is inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ so that $Q$ is at $(0,b),$ and $\overline{PR}$ is parallel to the $x$-axis, as shown below. Also, foci $F_1$ and $F_2$ lie on sides $\overline{QR}$ and $\overline{PQ},$ respectively. Find $\frac{PQ}{F_1 F_2}.$
Since it is a ratio question you can use a convenient ellipse.
Let X be the centre and M be the midpoint of PR
triangle QXF2 and triangle QMP are similar tiranges and they are both 30, 60 90 degree triangles.
So their sides are in the ration 1:2:sqrt3
We can let the focal length (c) to be 1 and we can Q be (0,1). Then a = 2.
From here, with routine calculations, you can work out that PQ/F_1 F_2 = 13/8.
+128448 For convenience let Q = (0, 1)
Let the center lie at (0,0)
Note that in the equation of the ellipse, "b" = 1
Then 1/2 the focal length = 1/sqrt (3)
So F1F2 = 2/sqrt (3) = (2/3)sqrt (3)
We can find "a" as sqrt [ 1^2 + (1/sqrt(3))^2 ] = sqrt (1 + 1/3) = sqrt [ 4/3 ] = 2/sqrt (3)
a^2 = 4/3
A line from Q through R will have a slope of tan (-60) = -sqrt (3)
So.....the equation of this line is y = -sqrt (3) + 1 (1)
To find the positive x values of R, sub (1) into the equation of the ellipse
x^2 / ( 4/3) + [ -sqrt (3)x + 1]^2 = 1
(3/4)x^2 + 3x^2 -2sqrt (3)x + 1 = 1
(15/4)x^2 - 2sqrt (3) x = 0
x ( 15/4x - 2sqrt (3) ) = 0
The second factor set to 0 will give ys the x coordinate of R
(15/4)x - 2sqrt (3) = 0
x = (4/15)2sqrt (3)
x = (8/15)sqrt (3)
And the y coordinate of R = -sqrt (3) (8/15)sqrt (3) + 1 = -24/15 + 1 = -9/15 = -3/5
And QR = PQ = sqrt [ (8/15sqrt (3)^2 + (1 - -3/5)^2 ] =
sqrt[ (8/5)^2 ( 1/3 + 1 ] = (8/5)sqrt [ 1 +1/3 ] = (8/5) sqrt ( 4/3) = (16/5) sqrt (1/3) = (16sqrt (3) / 15) =
(16/15)sqrt (3)
So
PQ / F1F2 = (16/15) sqrt (3) / [ (2/3)sqrt (3) ] = (16/15) (3/2) = 8/5
