Question

$A$ and $B$ being the fixed points $(a,0)$ and $(-a,0)$ respectively, obtain the equations giving the locus of $P$, when
$PA+PB=c$, a constant quantity.

Answer 1

Let the point $P$ be $(h,k)$
$PA=\sqrt{{(h-a)}^2+{(k-0)}^2}PB=\sqrt{{(h+a)}^2+{(k-0)}^2}$

$PA+PB=cPA=c-PB$

Squaring both sides

${PA}^2=c^2+{PB}^2-2cPB{(h-a)}^2+{\left(k\right)}^2=c^2{+(h+a)}^2+{\left(k\right)}^2-2cPB{h}^2+a^2-2ah+k^2=c^2+h^2+a^2+2ah+k^2-2cPB-4ah-c^2=-2cPB$

Again squaring both sides

${(4ah+c^2)}^2=4c^2{PB}^{2}16a^2{h}^2+c^4+8a{c}^{2}h=4c^2(h^2+a^2+2ah+k^2)16a^2{h}^2-4c^2{h}^2+8a{c}^{2}h-8c^{2}ah-4c^2{k}^2+c^4-4c^2{a}^2=04h^2(4a^2-c^2)-4c^2{k}^2=c^2(4a^2-c^2)$

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

$4x^2(4a^2-c^2)-4c^2{y}^2=c^2(4a^2-c^2)$

is the required equation of locus.