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

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the vertices, foci, and asymptote of 16x^2 - 25y^2=400      Log On


   

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

Hi, 

16x^2 - 25y^2=400

x%5E2%2F5%5E2+-+y%5E2%2F4%5E2=1

Note:Standard Form of an Equation of an Hyperbola is  %28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 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 (±sqrt%2841%29,0)