SOLUTION: Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (±2, 0); foci: (±3, 0)

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Question 1176575: Find the standard form of the equation of the hyperbola with the given characteristics.
Vertices: (±2, 0); foci: (±3, 0)

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. %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1

Foci is given to be, (3, 0) and (-3, 0), tells us that the hyperbola is the horizontal transverse type with the equation in item 1


%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1.... since the center is half way between foci, so it is at origin; so, h=0 and k=0

x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1
The distance to the focus is linear eccentricity c.
The point (3, 0) shows us the value of c=3:
The distance to the Vertices is a.
The point (2, 0) shows us the value of a=2

c%5E2=a%5E2%2Bb%5E2
3%5E2=2%5E2%2Bb%5E2
3%5E2-2%5E2=b%5E2
9-4=b%5E2
b%5E2=5
b=sqrt%285%29

then your equation is:
x%5E2%2F4-y%5E2%2F5=1