Question

Show the locus of the poles of tangents to the parabola $y^2=4ax$ with respect to the parabola $y^2=4bx$ is the parabola
$y^2=\frac{4b^2}{a}x$

Answer 1

The tangent to the parabola $y^2=4ax$ is $ty=x+a{t}^2$

Applying $T=0$ for the parabola $y^2=4bx$, we have $yk=2bx+2bh$ where $(h,k)$ is the pole of the tangent.

Comparing the two equations, we have

$\frac{t}{k}=\frac{1}{2b}=\frac{a{t}^2}{2bh}$

$\therefore t=\frac{k}{2b},t^2=\frac{h}{a}$

$\Rightarrow {\left(\frac{k}{2b}\right)}^2=\frac{h}{a}$

$\therefore k^2=\frac{4b^2}{a}h$ becomes the required locus.