Answer:
C. y squared/4 minus x squared/16 equals 1
Step-by-step explanation:
Given that vertices of hyperbola are (0,-2) and (0,2) and
Co-vertices of hyperbola as (-4,0) and (4,0) then a sketch will show you that this hyperbola has its transverse axis on the y-axis and has its center at the origin (0,0)
This means that the standard equation for the hyperbola should follow;
[tex]\frac{y^2}{a^2} -\frac{x^2}{b^2} =1[/tex]
Using the coordinates of the vertices, you can find the length of the transverse axis and the value of a as;
(0,-2) and (0,2) , 2--2=4
2a=4
a=4/2=2
a=2
Using the coordinates of the co-vertices, you can find the length of the conjugate axis as;
(-4,0) and (4,0) ⇒ 4--4=8
2b=8
b=8/2=4
b=4
substitute values in equation as;
[tex]\frac{y^2}{4} -\frac{x^2}{16} =1[/tex]