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
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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
{{{ graph( 300, 300, -5, 5, -10, 10,((17x^2/16-1)(256/17))^.5,-((17x^2/16-1)(256/17))^.5,4x,-4x) }}}