Question

HURRY IM TIMED

Answer 1

Centered at the origin: C=(0,0)=(h,k)→h=0, k=0
Focus: F=(0, 4)
Vertex: V=(0, sqrt 12)

The focus and the vertex are on a vertical line: x=0, then the equation has the form:
y^2/a^2-x^2/b^2=1

Distance between the Center and the Vertex: a=sqrt(12)
a^2=[sqrt(12)]^2→a^2=12

Distance between the Center and the Focus: c=4
c^2=(4)^2→c^2=16

c^2=a^2+b^2
Replacing c^2=16 and a^2=12
16=12+b^2
Solving for b^2: 
16-12=12+b^2-12
4=b^2
b^2=4

Replacing a^2=12 and b^2=4 in the equation of the hyperbola:
y^2/a^2-x^2/b^2=1
y^2/12-x^2/4=1

Answer: The equation of the hyperbola is y^2/12-x^2/4=1