Answer
$\frac{x^2}{4}-\frac{y^2}{6}=1$
Work Step by Step
Foci: $(±c,0)=(±\sqrt {10},0)$
$c=\sqrt {10}$
$c^2=a^2+b^2$
$a^2+b^2=10$
Use the point $(4,\sqrt {18})$:
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{16}{a^2}-\frac{18}{b^2}=1$
$\frac{16}{a^2}=\frac{18}{b^2}+1$
$\frac{16}{a^2}=\frac{18+b^2}{b^2}$
$\frac{a^2}{16}=\frac{b^2}{b^2+18}$
$a^2=\frac{16b^2}{b^2+18}$
$a^2+b^2=\frac{16b^2}{b^2+18}+b^2$
$10=\frac{16b^2}{b^2+18}+b^2~~$ (Multiply both sides by $b^2+18$)
$10b^2+180=16b^2+b^4+18b^2$
$0=b^4+24b^2-180$
$0=(b^2+30)(b^2-6)$
$b^2+30=0$ has no real solution.
$b^2-6=0$
$b^2=6$
$a^2+b^2=10$
$a^2=4$
Finally:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{4}-\frac{y^2}{6}=1$
