SOLUTION: Write an equation for the hyperbola that has eccentricity 3, center at (0,0), and vertex at (0,8) An equation for the hyperbola is...?

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Question 1149990: Write an equation for the hyperbola that has eccentricity 3, center at (0,0), and vertex at (0,8)
An equation for the hyperbola is...?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the hyperbola has a horizontal transverse axis and its standard form of the equation is:
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1
the hyperbola has a vertical transverse axis and its standard form of the equation is:
%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1

(h,k) is the center
a+is the distance between center+and vertex, the length of the transverse axis is 2a
the length of the conjugate axis is 2b
c is the distance between center and focus
c%5E2=a%5E2%2Bb%5E2
The foci of the hyperbola is (h,k ± c).
The vertices of the hyperbola is (h,k+± a).
Asymptotes of the hyperbola are: y-k= ±+%28a%2Fb%29%28x-h%29.
The formula for eccentricity+e is: e=sqrt%28a%5E2%2Bb%5E2%29%2Fa

given:

the hyperbola that has eccentricity 3, =>3=sqrt%28a%5E2%2Bb%5E2%29%2Fa
center at (0,0), =>h=0,k=0 and your formula becomes
y%5E2%2Fa%5E2-x%5E2%2Fb%5E2=1 (the standard equation for a hyperbola with a vertical transverse axis)
vertex at (0,8)=> a=8
then 3=sqrt%288%5E2%2Bb%5E2%29%2F8........solve for b%5E2
24=sqrt%288%5E2%2Bb%5E2%29.......square both sides
24%5E2=8%5E2%2Bb%5E2
b%5E2=24%5E2-8%5E2
b%5E2=576-64
b%5E2=512
and, your formula is: y%5E2%2F64-x%5E2%2F512=1