The equation is given to be:
[tex]x^2-3y^2+12=0[/tex]
We can write the equation in the standard form of the equation of a hyperbola to be:
[tex]\frac{\left(y-0\right)^2}{2^2}-\frac{\left(x-0\right)^2}{\left(2\sqrt{3}\right)^2}=1[/tex]
Therefore, we have the following parameters:
[tex]\left(h,\:k\right)=\left(0,\:0\right),\:a=2,\:b=2\sqrt{3}[/tex]
Recall the hyperbola foci definition:
[tex]\begin{gathered} \mathrm{For\:an\:up-down\:facing\:hyperbola,\:the\:Foci\:\left(focus\:points\right)\:are\:defined\:as}\:\left(h,\:k+c\right),\:\left(h,\:k-c\right),\: \\ \mathrm{where\:}c=\sqrt{a^2+b^2}\mathrm{\:is\:the\:distance\:from\:the\:center}\:\left(h,\:k\right)\:\mathrm{to\:a\:focus} \end{gathered}[/tex]
Therefore, the value of c will be:
[tex]\begin{gathered} c=\sqrt{2^2+(2\sqrt{3})^2} \\ c=4 \end{gathered}[/tex]
Therefore, the foci will be:
[tex]\begin{gathered} \left(h,\:k+c\right),\:\left(h,\:k-c\right)=\left(0,\:0+4\right),\:\left(0,\:0-4\right) \\ Foci=\left(0,\:4\right),\:\left(0,\:-4\right) \end{gathered}[/tex]
The correct option is the FIRST OPTION.
