Question 170285
Take note that ALL of the points given to you (both vertices and foci) all have a y-coordinate of 0. So this tells us that the hyperbola opens left and right like this:




<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/hyperbola.png" alt="Photobucket - Video and Image Hosting">




Take note that the distance from the center to either focus is 8 units. So let's call this distance "c" (ie {{{c=8}}})



Remember, the equation of any hyperbola opening left/right is 


{{{((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1}}}


So we need to find the values of h, k, a, and b



Now let's find the midpoint of the line connecting the vertices. This midpoint is the center of the hyperbola


x mid: Average the x-coordinates of the vertices: {{{(-8+8)/2 = 0/2 = 0}}}

So the x-coordinate of the center is 0. This means that h = 0

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y mid: Average the y-coordinates of the vertices: {{{(0+0)/2 = 0/2 = 0}}}
 
So the y-coordinate of the center is 0. This means that k = 0


So the center is (0,0) which means that h=0 and k=0


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Now because the hyperbola opens left and right, this means that the vertices are (h+a,k) and (h-a,k). In other words, you add and subtract the value of "a" to the x-coordinate of the center to get the vertices.



Since the value of "h" and "k" is 0, this means that the vertices become (0+a,0) and (0-a,0) then simplify to (a,0) and (-a,0)


So this tells us that a=6 and -a=-6 which simply means that a=6


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Now it turns out that the value of "b" is closely connected to the values of "a" and "c". They are connected by the equation



{{{a^2+b^2=c^2}}} 



{{{6^2+b^2=8^2}}} Plug in {{{a=6}}} and {{{c=8}}}



{{{36+b^2=64}}} Square 6 to get 36. Square 8 to get 64



{{{b^2=64-36}}} Subtract 36 from both sides.



{{{b^2=28}}} Subtract



{{{b=sqrt(28)}}} Take the square root of both sides. Note: only the positive square root is considered (since a negative distance doesn't make sense)



{{{b=2*sqrt(7)}}} Simplify the square root.



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Recap:


So we found the following: {{{h=0}}}, {{{k=0}}} (the x and y coordinates of the center), {{{a=6}}} and {{{b=2*sqrt(7)}}}




{{{((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1}}} Start with the general equation for a hyperbola (one that opens left/right)



{{{((x-0)^2)/(6^2)-((y-0)^2)/((2*sqrt(7))^2)=1}}} Plug in {{{h=0}}}, {{{k=0}}}, {{{a=6}}} and {{{b=2*sqrt(7)}}}



{{{((x-0)^2)/(36)-((y-0)^2)/(4*7)=1}}} Square 6 to get 36. Square {{{2*sqrt(7)}}} to get {{{4*7}}}



{{{((x-0)^2)/(36)-((y-0)^2)/(28)=1}}} Multiply



{{{(x^2)/(36)-(y^2)/(28)=1}}} Simplify



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Answer:



So the equation of the hyperbola that has the foci (8,0) and (-8,0) along with the vertices (-6,0) and (6,0) is:


{{{(x^2)/(36)-(y^2)/(28)=1}}}