SOLUTION: Find the equation in standard form of the hyperbola with an asymptote slope of 3/4 and foci (-2,-8) and (8,-8). a.(x-3)^2/9-(y+8)^2/16=1 b.(y-3)^2/16-(x+8)^2/9=1 c.(y+8)^2/9-(

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation in standard form of the hyperbola with an asymptote slope of 3/4 and foci (-2,-8) and (8,-8). a.(x-3)^2/9-(y+8)^2/16=1 b.(y-3)^2/16-(x+8)^2/9=1 c.(y+8)^2/9-(      Log On


   

Question 470819: Find the equation in standard form of the hyperbola with an asymptote slope of 3/4 and foci (-2,-8) and (8,-8).
a.(x-3)^2/9-(y+8)^2/16=1
b.(y-3)^2/16-(x+8)^2/9=1
c.(y+8)^2/9-(x-3)^2/9=1
d.(x-3)^2/16-(y+8)^2/9=1
e.(x+8)^2/9-(y-3)^2/16=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 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.

Asymptotes passing thru the center with slope = ± b/a 

foci being ± sqrt(a^2 + b^2)from center  along axis of symmetry y = k



asymptote slope b/a = 3/4   %28x-h%29%5E2%2F16+-+%28y-k%29%5E2%2F9+=+1 

foci (-2,-8) and (8,-8). c = sqrt(16+9) = 5, C(3,-8) %28x-3%29%5E2%2F16+-+%28y%2B8%29%5E2%2F9+=+1