SOLUTION: Find the equation of the hyperbola with center at the origin, foci at (-4,0) and (4,0), and eccentricity of 5/4.

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Question 1170472: Find the equation of the hyperbola with center at the origin, foci at (-4,0) and (4,0), and eccentricity of 5/4.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Foci at (-4,0) and (4,0) means the equation is of the general form

x%5E2%2Fa%5E2-y%5E2%2Fb%5E2+=+1

a is the distance from the center to each end of the transverse axis (i.e., to each vertex); b is the distance from the center to each end of the conjugate axis (i.e. to each co-vertex). a and b are related by

c%5E2+=+a%5E2%2Bb%5E2

where c is the distance from the center to each focus.

The eccentricity is c/a.

The given information is c=4 and eccentricity = 5/4.

(1) Use c and the eccentricity to find a.

(2) Use c%5E2+=+a%5E2%2Bb%5E2 to find b.

(3) Plug the a^2 and b^2 values into the general form of the equation.