SOLUTION: Find an equation of a hyperbola with vertices at (-9,4) and (-5,4) and asymptotes y=3x+25 and y=-3x-17.

Question 137050: Find an equation of a hyperbola with vertices at (-9,4) and (-5,4) and asymptotes y=3x+25 and y=-3x-17.

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
Find an equation of a hyperbola with vertices at (-9,4) and (-5,4) and asymptotes y=3x+25 and y=-3x-17.

First let's draw the asymptotes and the vertices:

We can tell that the hyperbola opens right and left
and has the equation

where the center is
The length of the semi-transverse axis is
The length of the semi-conjugate axis is
The two asymptotes have slopes � and pass
through the center (h,k).
The center is the midpoint between the vertices,
(-9,4) and (-5,4) which is the point (-7,4)
So (h,k) = (-7,4)
The transverse axis is the distance between the vertices
and which is units.
So the semi-transverse axis is half of that or .
That is
Since the asymptotes are and
We compare them to and find that their slopes
are �,
Since the slope are �, then
and since , we substitute that:
or
So the equation

becomes

Now we draw the defining rectangle which has base
2a and height 2b and has the center as its center.

Now we can sketch in the hyperbola:

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