SOLUTION: Find the standard form of the equation of the hyperbola with the given characteristics. asymptotes:y=+-4x a.(y^2)/(16/17)-(x^2)/(256/17)=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the standard form of the equation of the hyperbola with the given characteristics. asymptotes:y=+-4x a.(y^2)/(16/17)-(x^2)/(256/17)=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1       Log On


   

Question 470818: Find the standard form of the equation of the hyperbola with the given characteristics.
asymptotes:y=+-4x
a.(y^2)/(16/17)-(x^2)/(256/17)=1
b.(x^2)/(16/17)-(y^2)/(256/17)=1
c.(y^2)/(16)-(x^2)/(16)=1
d.(x^2)/(16)-(y^2)/(16)=1
e.(x^2)/(256/17)-(y^2)/(16/17)=1
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard form of the equation of the hyperbola with the given characteristics.
asymptotes:y=+-4x
a.(y^2)/(16/17)-(x^2)/(256/17)=1
b.(x^2)/(16/17)-(y^2)/(256/17)=1
c.(y^2)/(16)-(x^2)/(16)=1
d.(x^2)/(16)-(y^2)/(16)=1
e.(x^2)/(256/17)-(y^2)/(16/17)=1
**
Standard form for hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1
b.(x^2)/(16/17)-(y^2)/(256/17)=1
This is a hyperbola with horizontal transverse axis. (opens sideways)
Center: (0,0)
a^2=16/17
a=4/√17
b^2=256/17
b=16/√17
Slope=b/a=(16/√17)/(4√17)=4 (matches given slope=±4)
asymptotes:
y=±4x
ans:Equation b. is the correct choice
see graph below as a visual check on the answer.
..
y=±((17x^2/16-1)(256/17))^.5