Respuesta :
Answer:
D.
[tex]\frac{y^2}{12}-\frac{x^2}{6}=1[/tex]
Step-by-step explanation:
we are given
we can use standard equation of hyperbola
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1[/tex]
where
center=(h,k)
center at origin
so, h=0 and k=0
vertex is
[tex](0,\sqrt{12})[/tex]
we can use formula
vertices: (h, k + a)
we get
[tex]k+a=\sqrt{12}[/tex]
we can plug k=0
[tex]a=\sqrt{12}[/tex]
now, we can plug these values
[tex]\frac{(y-0)^2}{(\sqrt{12})^2}-\frac{(x-0)^2}{b^2}=1[/tex]
now, we are given it passes through [tex](2\sqrt{3} ,6)[/tex]
so, we have
[tex]x=2\sqrt{3},y=6[/tex]
we can plug these values and then we can solve for b
[tex]\frac{(6-0)^2}{(\sqrt{12})^2}-\frac{(2\sqrt{3}-0)^2}{b^2}=1[/tex]
and we get
[tex]36b^2-144=12b^2[/tex]
we can solve for b
and we get
[tex]b=\sqrt{6}[/tex]
now, we can plug these values
[tex]\frac{(y-0)^2}{(\sqrt{12})^2}-\frac{(x-0)^2}{(\sqrt{6})^2}=1[/tex]
we can simplify it
and we get
[tex]\frac{y^2}{12}-\frac{x^2}{6}=1[/tex]
