SOLUTION: Find the standard form of the equation of the hyperbola with the given characteristics. Foci: (±10, 0); asymptotes: y = ±4/3x

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Question 1176576: Find the standard form of the equation of the hyperbola with the given characteristics.
Foci: (±10, 0); asymptotes: y = ±4/3x
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

There are two standard forms:
1. %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1
2. = ±+m%28x-h%29%2Bk
Therefore, the equation, y = ±%284%2F3%29x tell us that h=0 and k=0
The center is the point (0,0).
Foci is given to be, (±10, 0), tells us that the hyperbola is the horizontal transverse type with the equation in item 1
%28x-0%29%5E2%2Fa%5E2-%28y-0%29%5E2%2Fb%5E2=1
x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1
The distance to the focus is linear eccentricity c.
The point (10, 0) shows us the value of c=10:
c%5E2=a%5E2%2Bb%5E2
y+= ±%284%2F3%29x shows us that
->b%2Fa=4%2F3
b=4a%2F3
10%5E2=a%5E2%2B%284a%2F3%29%5E2
100=a%5E2%2B16a%5E2%2F9
100=25a%5E2%2F9
900%2F25=a%5E2
36=a%5E2
a=6
then
b%2F6=4%2F3
b=24%2F3
b=8


x%5E2%2F6%5E2-y%5E2%2F8%5E2=1
and your equation in standard form is:
x%5E2%2F36-y%5E2%2F64=1