Question

$A$ and $B$ being the fixed points $(a,0)$ and $(-a,0)$ respectively, obtain the equations giving the locus of $P$, when $P{A}^2-P{B}^2=$ a constant quantity $=2k^2$

Answer 1

$A(a,0)$ $B(-a,0)$

Let the point $P$ be $(p,q)$
$PA=\sqrt{{(p-a)}^2+{(q-0)}^2}P{A}^2={(p-a)}^2+{\left(q\right)}^{2}PB=\sqrt{{(p-a)}^2+{(q-0)}^2}P{B}^2={(p+a)}^2+{\left(q\right)}^{2}P{A}^2-P{B}^2={(p-a)}^2+{\left(q\right)}^2-{\left(\right(p+a)}^2+{\left(q\right)}^2)2k^2=p^2+a^2-2ap+q^2-p^2-a^2-2ap-q^2-4ap=2k^{2}2ap+k^2=0$

Replacing $p$ by $x$

$2ax+k^2=0$

is the required equation of locus.