Question

$A$ and $B$ being the fixed points $(a,0)$ and $(-a,0)$ respectively, obtain the equations giving the locus of $P$, when $P{B}^2+P{C}^2=2P{A}^2$, $C$ being the point $(c,0)$.

Answer 1

$PA=\sqrt{{(h-a)}^2+{(k-0)}^2}PB=\sqrt{{(h+a)}^2+{(k-0)}^2}PC=\sqrt{{(h+c)}^2+{(k-0)}^2}$

$PA=\sqrt{{(h-a)}^2+{(k-0)}^2}PB=\sqrt{{(h+a)}^2+{(k-0)}^2}PC=\sqrt{{(h-c)}^2+{(k-0)}^2}{PB}^2+{PC}^2=2{PA}^2{(h+a)}^2+{\left(k\right)}^2+{(h-c)}^2+{\left(k\right)}^2=2({(h-a)}^2+{\left(k\right)}^2)h^2+a^2+2ah+k^2+h^2+c^2-2ch+k^2=2(h^2+a^2-2ah+k^2)h^2+a^2+2ah+k^2+h^2+c^2+2ch+k^2={2h}^2+2a^2-4ah+2k^2-2ch+6ah+c^2-a^2=0(6a-2c)h=a^2-c^2$

Replacing $h$ by $x$ and $k$ by $y$

$(6a-2c)x=a^2-c^2$

is the required locus