Write an equation of a hyperbola with vertices (3.-2) and (-9,-2) and foci (7,-2) and (-13,-2)
a.(x-3)^2/36-(y-2)^2/64=1
b.(x-3)^2/12-(y-2)^2/16=1
c.(x+3)^2/12-(y+2)^2/16=1
d.(x+3)^2/36-(y+2)^2/64=1

Question

Write an equation of a hyperbola with vertices (3.-2) and (-9,-2) and foci (7,-2) and (-13,-2) 
a.(x-3)^2/36-(y-2)^2/64=1 
b.(x-3)^2/12-(y-2)^2/16=1
c.(x+3)^2/12-(y+2)^2/16=1
d.(x+3)^2/36-(y+2)^2/64=1

anonymous · Answer

with the given vertices, can you find the center of the hyperbola?

anonymous · Answer

I dont know how to do this.  Teacher was explaining it but I didnt understand

anonymous · Answer

the midpoint of the two vertices is the center of your hyperbola...

anonymous · Answer

at least find that so you can eliminate some choices...

anonymous · Answer

$$
\frac { (x-h)^2} {a^2}- \frac {(y-k)^2}{b^2} =1
$$

(h,k) is the center. 
2a = the distance between the two vertices 
2 c=the distance between the two foci.

$$ b^2 = c^2 -a^2
$$

anonymous · Answer

Write an equation of a hyperbola with vertices (3.-2) and (-9,-2) and foci (7,-2) and (-13,-2) 
center=( -3, -2)

anonymous · Answer

2a =12 
a=6

anonymous · Answer

2c = 20
c=10

anonymous · Answer

$$
b^2 = 100 -36 =64
$$

anonymous · Answer

$$\frac {(x+3)^2}{36}-\frac {(y+2)^2}{64}=1
$$

anonymous · Answer

It is d.

anonymous · Answer

Thanks man.!  Thats a bunch of steps thanks for being patient with me.

anonymous · Answer

yw