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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?

waffles  Nov 20, 2017
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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

 

 

cool cool cool

CPhill  Nov 21, 2017

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