Question

Prove that the locus of the pole of a given straight line with respect to a series of confocals is a straight line which is the normal to that confocal which the straight line touches.

Answer 1

Let the pole of the line $lx+my=\mathrm{1....}\left(1\right)$
w.r.t. the confocal conic;-
$\frac{x^2}{a^2+\lambda }+\frac{y^2}{b^2+\lambda }=1$ be $(h,k)$
Then the equation of polar of $(h,k)$ w.r.t. the conic is
$\frac{xh}{a^2+\lambda }+\frac{ky}{b^2+\lambda }=\mathrm{1.....}\left(2\right)$
Since the two lines $\left(1\right)$ and $\left(2\right)$ are identical, so on comparing, we get
$\frac{h}{(a^2+\lambda )l}=\frac{k}{(b^2+\lambda )m}=1$
which gives $h=l(a^2+\lambda ),k=(b^2+\lambda )m$
Eliminating $\lambda$ between these two equations we get required locus, as
$\frac{x}{l}-\frac{y}{m}=a^2-b^2$
The above line is clearly perpendicular to $\left(1\right)$
Also the point of contact of this line with a confocal conic clearly lies on the locus. As it is the pole of the line with respect to that focal.