jmc

Question

An equilateral triangle PQR is inscribed in the ellipse x^2a^2 + y^2b^2 = 1, so that Q is at (0,b), and PR is parallel to the x-axis, as shown below. Also, foci F_1 and F_2 lie on sides QR and PQ, respectively. Find PQF_1 F_2. FIGURE

Accepted Answer

Without loss of generality, assume that

F_{1} F_{2} = 2,

so

c = 1.

Since triangle

Q F_{1} F_{2}

is equilateral,

b = 3

and

a = 2.

Let

s

be the side length of equilateral triangle

P QR,

and let

M

be the midpoint of

\overline{P R} .

Then

R M = \frac{s}{2} .

Also,

R F_{1} = QR - Q F_{1} = s - 2,

so the distance from

R

to the

x

-axis is

\frac{3}{2} (s - 2) .

Hence,

R = (\frac{s}{2}, - \frac{3}{2} (s - 2)) .

Substituting these coordinates into the equation of the ellipse, we get

\frac{( \frac{s}{2} ) ^{2}}{4} + \frac{( - \frac{3}{2} ( s - 2 ) ) ^{2}}{3} = 1.

This simplifies to

5 s^{2} = 16 s,

so

s = \frac{16}{5} .

Therefore,

\frac{P Q}{F _{1} F _{2}} = \frac{16/5}{2} = \frac{8}{5} .