The vertices of a hyperbola are at $(0,0)$ and $(10,0)$ and one of its foci is at $(18,0)$. The equation of the hyperbola is

The correct option is B $\frac{(x-5{)}^2}{25}-\frac{y^2}{144}=1$
Distance between vertices is $=2a=10-0=10\Rightarrow a=5$
Distance between vertex and focus $=ae-a=8$
$\Rightarrow e=1+\frac{8}{5}=\frac{13}{5}$
$\therefore b=5\times \sqrt{\frac{(13{)}^2}{5^2}-1}=5\times \frac{12}{5}=12$
Centre of hyperbola (midpoint of the line joining the two foci) $=(5,0)$
$\therefore \frac{(x-5{)}^2}{25}-\frac{y^2}{144}=1$