Answer: y^2/16-x^2/256=1
Solution:
The vertices are (0,±4): (0,-4) and (0,4)
The parabola has a vertical axis (the line that joins the vertices).
The center C=(h-k) is the midpoint between the vertices, then:
h=(0+0)/2→h=0/2→h=0
k=(-4+4)/2→k=0/2→k=0
C=(h,k)→C=(0,0)
The equation of the parabola has the form:
y^2/a^2-x^2/b^2=1
The distance between the vertices and the center is a, then a=4
y^2/4^2-x^2/b^2=1
y^2/16-x^2/b^2=1 (1)
The asympotes of this parabola are:
y=±(a/b)x
The given asympotes are: y=±(1/4)x, then:
a/b=1/4; a=4
4/b=1/4
Solving for b. Cross multiplication:
(4)(4)=(1)(b)
16=b
b=16
Replacing in the equation (1)
(1) y^2/16-x^2/16^2=1
y^2/16-x^2/256=1
Equation of the hyperbola:
y^2-(x/4)^2=1
y^2-x^2/4^2=1
y^2-x^2/16=1