Find the equation of the parabola whose focus is (3, 5) vertex is (1, 3).

Given S= (3, 5) and A =(1, 3)
Let $\overrightarrow{\mathrm{SA}}$ meet the directrix in Z(h. k)
Mid point of SZ= A(1,3) $\Rightarrow$$(\frac{\mathrm{h}+3}{2}, \frac{\mathrm{k}+5}{2})$ = (1,3) $\Rightarrow$Z(h,k) = (-1,1)
Slope of SZ = $\frac{5-1}{3+1} \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{directrix} = -1$
$\therefore$Equation to the directrix passing through Z(-1, 1) is
y-1 = -1(x+ 1) $\Rightarrow$ x+y = 0
Let P(x₁,y₁) $\in$ parabola and PM perpendicular to the directrix.
Then ${\mathrm{SP}}^2={\mathrm{PM}}^2\Rightarrow {\left({\mathrm{x}}_1-3\right)}^2+{\left({\mathrm{y}}_1-5\right)}^2={\left[\frac{{\mathrm{x}}_1+{\mathrm{y}}_1}{\sqrt{1+1}}\right]}^2$
$\begin{array}{l}\Rightarrow 2\left({\mathrm{x}}_1^2-6{\mathrm{x}}_1+9+{\mathrm{y}}_1^2-10{\mathrm{y}}_1+25\right)={\mathrm{x}}_1^2+2{\mathrm{x}}_1{\mathrm{y}}_1+{\mathrm{y}}_1^2\\ {\mathrm{x}}_1^2-2{\mathrm{x}}_1{\mathrm{y}}_1+{\mathrm{y}}_1^2-12{\mathrm{x}}_1-20{\mathrm{y}}_1+68=0\end{array}$
$\therefore$Equation of the locus of P is the parabola ${\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2-12\mathrm{x}-20\mathrm{y}+68=0$