SOLUTION: Please help me with this question. Find an equation for the hyperbola that satisfies the given conditions: 12. Vertices (0, ±4), asymptotes y = ± 1/2x

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me with this question. Find an equation for the hyperbola that satisfies the given conditions: 12. Vertices (0, ±4), asymptotes y = ± 1/2x      Log On


   

Question 1049488: Please help me with this question.


Find an equation for the hyperbola that satisfies the given conditions:
12. Vertices (0, ±4), asymptotes y = ± 1/2x
Found 2 solutions by ewatrrr, advanced_Learner:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Vertices (0, ±4), asymptotes y = ± 1/2x
OPens Up an Down
Standard Form of an Equation of an Hyperbola opening up and down is:
%28y-k%29%5E2%2Fb%5E2+-+%28x-h%29%5E2%2Fa%5E2+=+1 with C(h,k) and vertices 'b' units up and down from center, 2b the length of the transverse axis
Foci sqrt%28a%5E2%2Bb%5E2%29units units up and down from center, along x = h
& Asymptotes Lines passing thru C(h,k), with slopes m = ± b/a
Vertices (0, ±4) therefore C(0,0)
%28y%29%5E2%2F4%5E2+-+%28x%29%5E2%2Fa%5E2+=+1
asymptotes y = ± 1/2x
m = ± b/a
4/a = 1/2
a = 8
%28y%29%5E2%2F4%5E2+-+%28x%29%5E2%2F8%5E2+=+1

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
the equation is
centre is at (0,0)
since vertices(0,4) and V(0,-4) a=4
find b from the asymptotes b=2





the answer is
y%5E2%2F16-x%5E2%2F4=1