SOLUTION: What is the standard form of a hyperbola with the given conditions foci at (-13,0) (13,0) vertices at (12,0) (-12,0)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the standard form of a hyperbola with the given conditions foci at (-13,0) (13,0) vertices at (12,0) (-12,0)      Log On


   

Question 947817: What is the standard form of a hyperbola with the given conditions
foci at (-13,0) (13,0)
vertices at (12,0) (-12,0)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The transverse axis (the segment connecting the vertices) is a segment of the x-axis,
so the transverse axis of this hyperbola is "horizontal" (in the same direction as the x-axis).
The center of the hyperbola is the midpoint of the transverse axis,
and the midpoint of the segment connecting (12,0) and (-12,0) is (0,0), the origin,
so this hyperbola is centered at the origin, with a horizontal transverse axis.
A hyperbola centered at the origin, with a horizontal transverse axis has an equation of the form
x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1 , with some positive numbers a and b .
With that equation, we get
x%5E2%2Fa%5E2%3E=1 --> x%5E2%3E=a%5E2 --> system%28x%3C=-a%2C%22or%22%2Cx%3E=a%29 ,
meaning that the points %22%28%22-a%22%2C%220%22%29%22 and %22%28%22a%22%2C%220%22%29%22 are the vertices,
so we know that highlight%28a=12%29 .
We also know that the foci are at a distance c from the center,
and that a%5E2%2Bb%5E2=c%5E2 .
In the case of this hyperbola, obviously c=13 ,
and knowing that a=12 ,
we can substitute the values for a and c into a%5E2%2Bb%5E2=c%5E2 ,
to get the equation 12%5E2%2Bb%5E2=13%5E2 , which will give us b .
12%5E2%2Bb%5E2=13%5E2-->144%2Bb%5E2=169-->b%5E2=169-144-->b%5E2=25-->b%5E2=25
highlight%28b=5%29 .
With values for a and b ,
we can substitute into x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1 ,
to get the equation for our hyperbola as
x%5E2%2F12%5E2-y%5E2%2F5%5E2=1 or highlight%28x%5E2%2F144-y%5E2%2F25=1%29