SOLUTION: 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 2

Algebra ->  Graphs -> SOLUTION: 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 2      Log On


   



Question 1077688: 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?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The ellipse has
a=50%2F2=25 (the semi-major axis), and
b=40%2F2=20 (the semi-minor axis).
The focal distance, c , can be found from
b%5E2%2Bc%5E2=a%5E2 .
Substituting known values, 20%5E2%2Bc%5E2=25%5E2-->c%5E2=625-400=225--> c=15 .
So, it looks like this (maybe shifted and/or rotated).
According to the definition of ellipse,
if P is a point of an ellipse with foci A and B,
PA%2BPB=2a=50 .

The hyperbola has
c=15 , just like the ellipse,
and has b=20%2F2=10 .
Knowing that in a hyperbola
a%5E2%2Bb%5E2=c%5E2 , we can find a .
a%5E2%2B15%5E2=10%5E2
a%5E2%2B225=100
a%5E2=225-100=125
a=sqrt%28125%29
In a hyperbola with foci A and B, for any point P, by definition
abs%28PA-PB%29=2a=2sqrt%28125%29 .

Squaring both sides in the equations found involving the distances PA and PB,
%28PA%2BPB%29%5E2=2500-->PA%5E2%2BPB%5E2%2B2%2APA%2APB=2500 and
%28PA-PB%29%5E2=500--->PA%5E2%2BPB%5E2-2%2APA%2APB=500 .
Subtracting PA%5E2%2BPB%5E2-2%2APA%2APB=500 from PA%5E2%2BPB%5E2%2B2%2APA%2APB=2500 , we get
PA%5E2%2BPB%5E2%2B2%2APA%2APB-PA%5E2-PB%5E2%2B2%2APA%2APB=2500-500 --> 4%2APA%2APB=2000 --> PA%2APB=2000%2F4=highlight%28500%29