SOLUTION: what is the center, foci, and asymptotes for the conic? x^2/16-y^2/25=1

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Question 438565: what is the center, foci, and asymptotes for the conic?
x^2/16-y^2/25=1
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

Standard Form of an Equation of an Hyperbola opeing right and left is  

            %28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 

where Pt(h,k) is a center  with vertices 'a' units right and left of center.

x^2/16-y^2/25=1  C(0,0) with vertices V(-4,0) and V(4,0) - the x-intercepts

Foci(c=sqrt%2841%29: F(-6.4,0) and F(6.4,0)

Asymptotes: y = ± (5/4)x