SOLUTION: What is the equation of the hyperbola with vertices (0,4) and (0,-4) and asymptotic y=(1/3)x and y=(-1/3x)

We plot the vertices and graph the asymptotes:



The center of the hyberbola is (0,0) and it opens up and down since its
vertices are one above the other, so its equation is of the form

  

and since the center (h,k) is half-way between the vertices, the center
is the origin and (h,k) = (0,0) and therefore the equation of the hyperbola
is of the form

  

The top and bottom of the defining rectangle for the hyperbola will
be horizontal lines through the two vertices drawn between the two 
asymptotes.  These will be the horizontal lines y=4 and y=-4. So we 
draw those two horizontal lines through the vertices:



Next we determine where those horizontal lines intersect the asymptotes
so that we can complete the defining rectangle:

We solve the system  by substitution

12 = x

So the upper right corner of the defining rectangle is (12,4) and by
symmetry the other three corners are (-12,4),(-12,-4) and (12,-4).

So we draw in the sides of the defining rectangle, and sketch in the
hyperbola:



The transverse axis is the line between the two vertices, and the
semi-transverse axis is known as "a".  so a=4 (the distance between the
center and either vertex. 

The conjugate axis is the line through the center across the defining
rectangle, and the semi-conjugate axis is known as "b".  so b=12 
(the distance between the center and the middle of the side of the
defining rectangle.  Therefore the equation of the hyperbola becomes:

  

  

  

Clear of fractions by multiplying through by the LCD of 144

9y² - x² = 144

Edwin