The point $\mathrm{P}$ moves such that the sum of the squares of its distances from two fixed points $\mathrm{A}\left(\mathrm{a}, 0\right), \mathrm{B}\left(-\mathrm{a}, 0\right)$ is $8{\mathrm{a}}^2$ the locus of $\mathrm{P}$ is

let $\mathrm{P}({\mathrm{x}}_1, {\mathrm{y}}_1)$ be Locus of point.

Given points A=(a, 0), B=(-a, 0)

Given condition is ${\mathrm{PA}}^2+{\mathrm{PB}}^2=8{\mathrm{a}}^2$

$$\begin{gathered} \Rightarrow {({\mathrm{x}}_1-\mathrm{a})}^2+{{\mathrm{y}}_1}^2+{({\mathrm{x}}_1+\mathrm{a})}^2+{{\mathrm{y}}_1}^2=8{\mathrm{a}}^2 \\ \Rightarrow {{\mathrm{x}}_1}^2+{\mathrm{a}}^2-2{\mathrm{ax}}_1+{{\mathrm{y}}_1}^2+{{\mathrm{x}}_1}^2+{\mathrm{a}}^2+2{\mathrm{ax}}_1+{{\mathrm{y}}_1}^2=8{\mathrm{a}}^2 \\ \Rightarrow 2{{\mathrm{x}}_1}^2+2{{\mathrm{y}}_1}^2=6{\mathrm{a}}^2 \\ \Rightarrow {{\mathrm{x}}_1}^2+{{\mathrm{y}}_1}^2=3{\mathrm{a}}^2 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{P} \mathrm{is} {\mathrm{x}}^2+{\mathrm{y}}^2=3{\mathrm{a}}^2 \end{gathered}$$