SOLUTION: find the vertices, foci, and asymptote of 16x^2 - 25y^2=400

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Question 612564: find the vertices, foci, and asymptote of 16x^2 - 25y^2=400
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
16x^2 - 25y^2=400

Note:Standard Form of an Equation of an Hyperbola is opening right or left
where Pt(h,k) is a center with vertices 'a' units right and left of center. V are (±5,0)
Asymptotes passing thru the center with slope = ± b/a , y = ±4/5x
foci being ± sqrt(a^2 + b^2) from center along axis of symmetry y = b
foci are (±,0)


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