Question

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

Answer 1

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

Squaring both sides

${(h-a)}^2+{\left(k\right)}^2=n^2{(h+a)}^2+{\left(k\right)}^2{h}^2+a^2-2ah+k^2=n^2{(h}^2+a^2+2ah+k^2\left)\right(n^2-1)h^2+(n^2-1)k^2+(n^2-1)a^2+(n^2+1)2ah=0(n^2-1\left)\right(h^2+k^2+a^2)+(n^2+1)2ah=0$

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

$(n^2-1)(x^2+y^2+a^2)+(n^2+1)2ax=0$

is the required equation of locus.