A circle has the same center as an ellipse and passes through the foci \(F_1\) and \(F_2\) of the ellipse. The two curves intersect in 4 points. Let P be any point of intersection. If the major axis of the ellipse is 15 and the area of triangle \(PF_1 F_2\) is 26, compute the distance between the foci.

Guest Jul 31, 2019

#1**+2 **

Let the center = (0, 0)

Let the major axis be along the x axis

a = 7.5

c^2 = a^2 - b^2

c^2 = 7.5^2 - b^2 ⇒ b^2 = 7.5^2 - c^2

2c = the base of triangle PF_{1}F_{2}

Radius of circle = c

Area of triangle = (1/2) (2c) ( height) = 26 ⇒ c (height) = 26 ⇒ height = 26/c

So.....P is on the ellipse and circle

Let P = ( x, 26/c)

And the equation of the circle = x^2 + y^2 = c^2 ⇒ x^2 + (26/c)^2 = c^2 ⇒ x^2 = c^2 - (26/c)^2

Using the equation for an ellipse :

x^2/ a^2 + y^2 /b^2 = 1 and substituting, we have :

[c^2 - (26/c)^2] / 7.5^2 + (26/c)^2 / [ 7.5^2 - c^2] = 1

Solving this for c is a little difficult but we get that c ≈ 5.5

And the height of the triangle = 26/c = 26/5.5 = 52/11 = y coordinate of P

And the x ooordinate for P = sqrt (5.5^2 - (26/5.5)^2 ) ≈ 2.81

So P = (2.81, 52/11)

So...the distance between the foci 2c = 2(5.5) ≈ 11

Here's a pic :

CPhill Jul 31, 2019