Question

Two equal parabolas, A and B, have the same vertex and axis but have their concavities turned in opposite directions; prove that the locus of poles with respect to B of tangents to A is the parabola A.

Answer 1

Let the parabola A be $y^2=4ax$, while B be $y^2=-4ax$

Tangent to A is written as $ty=x+a{t}^2$

Poles w.r.t. B would be given by $yk=-2ax-2ah$

When compared this with the tangent equation, we have

$\frac{t}{k}=\frac{1}{-2a}=\frac{a{t}^2}{-2ah}$

$\Rightarrow t=\frac{-k}{2a}...\left(1\right)$

Also, $t^2=\frac{h}{a}...\left(2\right)$

Squaring and equating, we get

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

$\therefore \frac{k^2}{4a^2}=\frac{h}{a}$

i.e. $k^2=4ah$

Hence proved.