Answer
$\frac{y^2}{8}-x^2=1$
Work Step by Step
Foci: $(0,±c)=(0,±3)$
$c=3$
$c^2=a^2+b^2$
$a^2+b^2=9$
Use the point $(1,4)$:
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{4^2}{a^2}-\frac{1^2}{b^2}=1$
$\frac{16}{a^2}-\frac{1}{b^2}=1$
$\frac{16}{a^2}=\frac{1}{b^2}+1=\frac{1+b^2}{b^2}$
$\frac{a^2}{16}=\frac{b^2}{1+b^2}$
$a^2=\frac{16b^2}{b^2+1}$
$a^2+b^2=\frac{16b^2}{b^2+1}+b^2$
$9=\frac{16b^2}{b^2+1}+b^2~~$ (Multiply both sides by $b^2+1$)
$9b^2+9=16b^2+b^4+b^2$
$0=b^4+8b^2-9$
$0=(b^2-1)(b^2+9)$
$b^2-1=0$
$b^2=1$
$b^2+9=0$ has no real solution.
$a^2+b^2=9$
$a^2=9-1=8$
Finally:
$\frac{y^2}{8}-\frac{x^2}{1}=1$
$\frac{y^2}{8}-x^2=1$
