Answer:
The equation of this hyperbola is
[tex]\frac{(y-3)^2}{2^2} -\frac{(x+5)^2}{6^2} =1[/tex]
Step-by-step explanation:
Given from the question,
center of hyperbola (h,k) is (-5,3)
vertices of hyperbola are at (-5,-5) and ( -5,-1)
You know the length of the transverse is 2a, and can be found using the vertices given as -1--5=4
2a=4, hence a=4/2 =2
a=2
The coordinates of the co-vertice are (-11,-3) and (1,-3)
You know the length of the conjugate axis is 2b and can be found using the co-vertices given as 1--11=12
12=2b,
b=6
The standard equation of a hyperbola with center h,k and transverse axis parallel to the y-axis is
[tex]\frac{(y-k)^2}{a^2} -\frac{(x-h)^2}{b^2} =1[/tex]
substitute values as
[tex]\frac{(y-3)^2}{2^2} -\frac{(x+5)^2}{6^2} =1[/tex]
