Hello Joslen,

The vertices (0,±4) are on the y-axis, and the center of the hyperbola will be midpoint of the line segment joining the two vertices. Thus, the center is (0,0). The standard form of the equation of a hyperbola with center at the origin (0,0) and vertices (0,±b) is given by

**y**^{2}**/b**^{2}** - x**^{2}**/a**^{2}** = 1**

where a and b are positive constants. Since the vertices in your problem are (0,±4), we have **b = 4**. Also, for the equation in the form above, it can be shown that the **asymptotes of the hyperbola are given by**

** y = ±(b/a)x**

Since the asymptotes in your problem are the lines y = ±(2/3)x, we can write

**b/a = 2/3.**

Now substitute b = 4 and solve the above for a.

4/a = 2/3

2a = 12

**a = 6**

Therefore, the equation of the hyperbola is

y^{2}/4^{2} - x^{2}/6^{2} = 1, or

**y**^{2}**/16 - x**^{2}**/36 = 1**

Hope that helps! Let me know if you need any further explanation.

William