Find the equation of locus of point , the sum of whose distances from (0,2),(0,–2) is 6.

Let P(x, y) be any point on the locus.
Given that the sum of distance from P.
to A(0, 2), B(0, –2) is 6.
$\text{ i.e., }\mathrm{PA}+\mathrm{PB}=6\Rightarrow \mathrm{PA}=6-\mathrm{PB}$
S.O.B.S
$\begin{array}{l}\Rightarrow {\mathrm{PA}}^2=36+{\mathrm{PB}}^2-12\left(\mathrm{PB}\right)\\ \Rightarrow {\mathrm{PA}}^2-36-{\mathrm{PB}}^2=-12\left(\mathrm{PB}\right)\\ \Rightarrow (\mathrm{x}-0{)}^2+(\mathrm{y}-2{)}^2-36-\left[(\mathrm{x}-0{)}^2+(\mathrm{y}+2{)}^2\right] =-12\left(\mathrm{PB}\right)\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+4-4\mathrm{y}-36-{\mathrm{x}}^2-{\mathrm{y}}^2-4-4\mathrm{y}=-12\left(\mathrm{PB}\right)\\ \Rightarrow -8\mathrm{y}-36=-12\left(\mathrm{PB}\right)\Rightarrow 2\mathrm{y}+9=3\left(\mathrm{PB}\right)\end{array}$
Again, S.O.B.S
$\begin{array}{l}\Rightarrow 4{\mathrm{y}}^2+81+36\mathrm{y}=9(\mathrm{PB}{)}^2\\ \Rightarrow 4{\mathrm{y}}^2+81+36\mathrm{y}=9\left[(\mathrm{x}-0{)}^2+(\mathrm{y}+2{)}^2\right]\\ \Rightarrow 4{\mathrm{y}}^2+81+36\mathrm{y}=9\left[{\mathrm{x}}^2+{\mathrm{y}}^2+4+4\mathrm{y}\right]\\ \Rightarrow 4{\mathrm{y}}^2+81+36\mathrm{y}=9{\mathrm{x}}^2+9{\mathrm{y}}^2+36+36\mathrm{y}\\ \Rightarrow 9{\mathrm{x}}^2+9{\mathrm{y}}^2+36-4{\mathrm{y}}^2-81=0\\ \Rightarrow 9{\mathrm{x}}^2+5{\mathrm{y}}^2-45=0\Rightarrow \frac{{\mathrm{x}}^2}{5}+\frac{{\mathrm{y}}^2}{9}=1\end{array}$
$\therefore \text{ The required equation of locus is }\frac{{\mathrm{x}}^2}{5}+\frac{{\mathrm{y}}^2}{9}=1$