Find the equation of locus of P, if A = (2, 3), B = (2, –3) and PA + PB = 8.

Given points are A = (2, 3), B = (2, –3)
Let P(x, y) is any point on the locus
Given condition is PA + PB = 8
PA = 8 – PB
s.o.b.s
$\begin{array}{l}\Rightarrow (\mathrm{PA}{)}^2=(8-\mathrm{PB}{)}^2\\ \Rightarrow {\mathrm{PA}}^2=64+{\mathrm{PB}}^2-16\mathrm{PB}\end{array}$
$\begin{array}{l}\Rightarrow (\mathrm{x}-2{)}^2+(\mathrm{y}-3{)}^2=64+(\mathrm{x}-2{)}^2+(\mathrm{y}+3{)}^2-16\mathrm{PB}\\ \Rightarrow 16\mathrm{PB}=(\mathrm{y}+3{)}^2-(\mathrm{y}-3{)}^2+64\\ \Rightarrow 16\mathrm{PB}={\mathrm{y}}^2+9+6\mathrm{y}-{\mathrm{y}}^2-9+6\mathrm{y}+64\\ \Rightarrow 16\mathrm{PB}=12\mathrm{y}+64\\ \Rightarrow 16\mathrm{PB}=4(3\mathrm{y}+16)\Rightarrow 4\mathrm{PB}=3\mathrm{y}+16\end{array}$
$\text{ s.o.b.s; }\Rightarrow 16{\mathrm{PB}}^2=(3\mathrm{y}+16{)}^2$
$\begin{array}{l}\Rightarrow 16\left[(\mathrm{x}-2{)}^2+(\mathrm{y}+3{)}^2\right]=9{\mathrm{y}}^2+256+96\mathrm{y}\\ \Rightarrow 16\left[{\mathrm{x}}^2+4-4\mathrm{x}+{\mathrm{y}}^2+9+6\mathrm{y}\right]=9{\mathrm{y}}^2+256+96\mathrm{y}\\ \Rightarrow 16{\mathrm{x}}^2+64-64\mathrm{x}+16{\mathrm{y}}^2+144+96\mathrm{y}-9{\mathrm{y}}^2-256-96\mathrm{y}=0\\ \Rightarrow 16{\mathrm{x}}^2+7{\mathrm{y}}^2-64\mathrm{x}-48=0\end{array}$
$\text{ locus of }\mathrm{P}(\mathrm{x},\mathrm{y})\text{ is }16{\mathrm{x}}^2+7{\mathrm{y}}^2-64\mathrm{x}-48=0$