One focus of hyperbola located at (1, –3) and the corresponding directrix is  y=2. Find the equation of the hyperbola of its eccentricity is 3/2

Given $\mathrm{focus} \mathrm{S}=(1,-3),\mathrm{e}=\frac{3}{2}$
Equation of directrix is y – 2 = 0
Let p(x, y) be any point on hyperbola $\frac{\mathrm{SP}}{\mathrm{PM}}=\mathrm{e}$
$\begin{array}{l}\Rightarrow \mathrm{SP}=\mathrm{e}\left(\mathrm{PM}\right)\\ \Rightarrow \sqrt{(\mathrm{x}-1{)}^2+(\mathrm{y}+3{)}^2}=\frac{3}{2}\frac{|\mathrm{y}-2|}{\sqrt{0+1}}\\ \Rightarrow \sqrt{{\mathrm{x}}^2+1-2\mathrm{x}+{\mathrm{y}}^2+9+6\mathrm{y}}=\frac{3}{2}(\mathrm{y}-2)\end{array}$
S.O.B.S.0
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+6\mathrm{y}+10=\frac{9}{4}(\mathrm{y}-2{)}^2\\ 4{\mathrm{x}}^2+4{\mathrm{y}}^2-8\mathrm{x}+24\mathrm{y}+40=9\left({\mathrm{y}}^2+4-4\mathrm{y}\right)\\ 4{\mathrm{x}}^2+4{\mathrm{y}}^2-8\mathrm{x}+24\mathrm{y}+40-9{\mathrm{y}}^2-36+36\mathrm{y}=0\\ \therefore 4{\mathrm{x}}^2-5{\mathrm{y}}^2-8\mathrm{x}+60\mathrm{y}+4=0\end{array}$