An ellipse and a hyperbola have the same foci, A and B, and intersect at four points. The ellipse has major axis 50, and minor axis 40. The hyperbola has conjugate axis of length 20. Let P be a point on both the hyperbola and ellipse. What is PA* PB?
We can let the equation of the ellipse be
x^2 / 625 + y^2 / 400 = 1 (1)
And we can let the equation of the hyperbola be
x^2 / 100 - y^2 / 125 = 1 (2)
The focii of each are A = (15,0) and B = (-15,0)
Setting (1) = (2) we have that
x^2 / 100 - x^2/625 = y^2 / 125 + y^2 / 400
[ 525]x^2 / 62500 = [525] y^2 / 50000
21x^2 / 2500 = 21y^2 / 2000
x^2 = (5/4)y^2
(.8)x^2 = y^2
Using (1) to find the x coordinate of the intersection, we have that
x^2 / 625 + .8x^2/ 400 = 1
[ 400x^2 + 500x^2 ] = 250000
x^2 = 250000/ 900
x^2 = 2500 / 9
x = 50/3
So
(2500/9) = (5/4)y^2
(10000 / 45) = y^2
(2000/9) = y^2
(20/3)√5 = y
So one "P" is ( 50/3 , (20√5/ 3 )
So PA = √ [ (50/3 - 15)^2 + (2000/9) ] = 15
And PB = √ [ (50/3 + 15)^2 + (2000/9) ] = 35
So PA * PB = 15 * 35 = 525 = [ "a^2" of the ellipse - "a^2" of the hyperbola ]
Here's a graph, here : https://www.desmos.com/calculator/m8haudkxxi