Respuesta :
Answer:
[tex]\frac{y^2}{16}-\frac{x^2}{20} =1[/tex]
Step-by-step explanation:
We are given equation of hyperbola
Vertices are (0,-4) (0,4)
we can use formula
vertices : (h , k+a) and (h,k-a)
we can compare and find h , k and 'a'
[tex](h,k+a)=(0,4)[/tex]
[tex]h=0[/tex]
[tex]k+a=4[/tex]
[tex](h,k-a)=(0,-4)[/tex]
[tex]k-a=-4[/tex]
we can add both equations
we get
[tex]2k=0[/tex]
[tex]k=0[/tex]
now, we can find 'a'
[tex]0+a=4[/tex]
[tex]a=4[/tex]
now, we can use foci formula
foci: (h, k + c), (h, k - c)
[tex](h,k+c)=(0,6)[/tex]
[tex]c=6[/tex]
we know that
[tex]c^2=a^2+b^2[/tex]
so, we can solve for b
[tex]6^2=4^2+b^2[/tex]
[tex]b=2\sqrt{5}[/tex]
now, we can use standard equation of hyperbola
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2} =1[/tex]
now, we can plug values
[tex]\frac{(y-0)^2}{4^2}-\frac{(x-0)^2}{(2\sqrt{5})^2} =1[/tex]
[tex]\frac{y^2}{16}-\frac{x^2}{20} =1[/tex]
