SOLUTION: The question is: Determine the vertices, asymptotes, and foci of the hyperbola defined by 144x^2-81y^2=81 Answers are: (A) vertices (0, +1) asymptotes: y=+4/3x, foci (0, +5/4)

Algebra ->  College  -> Linear Algebra -> SOLUTION: The question is: Determine the vertices, asymptotes, and foci of the hyperbola defined by 144x^2-81y^2=81 Answers are: (A) vertices (0, +1) asymptotes: y=+4/3x, foci (0, +5/4)      Log On


   

Question 30781: The question is:
Determine the vertices, asymptotes, and foci of the hyperbola defined by 144x^2-81y^2=81
Answers are:
(A) vertices (0, +1) asymptotes: y=+4/3x, foci (0, +5/4)
(B) vertices (0, +1) asymptotes: y=+3/4x, foci: (0, +5/4)
(C) vertices (+3/4,0) asymptotes: y=+3/4x. foci: (+5/4,0)
(D) vertices (+3/4,0) asymptotes: y=+4/3s, foci: (+5/4,0)]
Thanks for your help with this.
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
(144x^2-81y^2=81)(1/81)
(16/9)(x^2) - y^2 = 1
(x^2)/(9/16) - y^2 = 1
horizontal transverse axis
a=3/4
b=1
a^2+b^=c^2
c=(5/4)
vertices: (3/4,0);(-3/4,0)
asymptote: y=(4/3 or -4/3)x
foci: (5/4,0);(-5/4,0)
*You are missing some terms*