For the hyperbola $9{\mathrm{x}}^2-16{\mathrm{y}}^2-18\mathrm{x}+32\mathrm{y}-151=0$

The given hyperbola can be written as

$\frac{(\mathrm{x}-1{)}^2}{16}-\frac{(\mathrm{y}-1{)}^2}{9}=1$

or $\frac{{\mathrm{X}}^2}{16}-\frac{{\mathrm{Y}}^2}{9}=1\text{ (where }\mathrm{X}=\mathrm{x}-1,\mathrm{Y}=\mathrm{y}-1\text{ ) }$

$\mathrm{e}=\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$

Directrices are$\mathrm{X}=\pm \mathrm{a}/\mathrm{e}.$ Therefore,

 $\mathrm{x}-1=\pm \frac{16}{5}$

or $\mathrm{x}=\frac{21}{5}\text{ and }\mathrm{x}=-\frac{11}{5}$

Length of latus rectum $=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{9}{2}$

The foci are given by

$\begin{array}{r}\mathrm{X}=\pm \mathrm{ae},\mathrm{Y}=0\\ \text{ or }(6,1),(-4,1)\end{array}$