Question

Find the locus of the foci of conics which have a common point and a common director circle.

Answer 1

Let the semi-axes of the conic be $(\alpha ,\beta )$
$C$ be the center of the director-circle,
$\sqrt{{\alpha }^2+{\beta }^2}$be its radius and $p$ be the common point.

Let $s$ and $s^{{}^{\prime }}$ be the foci, then
$sp.s^{{}^{\prime }}p={\alpha }^2+{\beta }^2-c{p}^2$
Take a point $p^{{}^{\prime }}$ on $pc$ produced such that $pc=c{p}^{{}^{\prime }}$; clearly.$s^{{}^{\prime }}p=s{p}^{{}^{\prime }}$. So
$s{p}^{{}^{\prime }}.s^{{}^{\prime }}p={\alpha }^2+{\beta }^2-c{p}^2$ .....(1)
Let the common point be $p(a,o)$.

Clearly $p^{{}^{\prime }}$ is $(-a,o)$ as c is the origin and $p{p}^{{}^{\prime }}$ passes through $c$.
If $(x,y)$ be the co-ordinates of the focus $3$, then by (1), we get
$[{(x+a)}^2+{(y-0)}^2][{(x-a)}^2+{(y-0)}^2]$
$={({\alpha }^2+{\beta }^2-a^2)}^2$
which is the required locus.