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 ${{y}^{2}}=\dfrac{4{{b}^{2}}}{a}x$.

Answer 1

Hint: We use the equation of tangent to the parabola $\left( y=mx+\dfrac{a}{m} \right)$ and equation of polar of the parabola $\left( ky=2a\left( x+h \right) \right)$.

Complete step-by-step answer:

Consider the above picture.
There are two parabolas green ${{y}^{2}}=4bx$ and orange ${{y}^{2}}=4ax$.
From point $A(h,k)$ on the green parabola, a black tangent to the orange parabola is drawn. The point where the tangent touches the orange parabola is $B$.
Since $AB$ is the tangent to the orange parabola, we can write its equation using the general equation of tangent, which is given as $y=mx+\left( \dfrac{a}{m} \right)$, where $~m$ is the slope of the tangent to the parabola.
So we can write the equation of line $AB$ as ,
$AB:y=mx+\left( \dfrac{a}{m} \right)$
Multiplying throughout by $m$ we get,
$AB:{{m}^{2}}x-my+a=0...(i)$
It is given that$A(h,k)$ is the pole of $AB$ with respect to parabola ${{y}^{2}}=4bx$.
We know that equation of polar of a general parabola ${{y}^{2}}=4ax$ with respect to a general point $\left( h,k \right)$ is $ky=2a\left( x+h \right)$.
So the equation of the polar for our parabola is
$ky=2b\left( x+h \right)$
$ky=2bx+2bh$
$2bx+2bh-ky=0...(ii)$
Since equations $\left( i \right)$ and $\left( ii \right)$ are representing the same line $AB$ their coefficients must be proportional.
So,
$\dfrac{2b}{{{m}^{2}}}=\left( \dfrac{-k}{-m} \right)=\dfrac{2bh}{a}$
Which gives,
$\dfrac{2b}{k}=m...(iii)$
$\dfrac{a}{h}={m}^{2}...(iv)$
Substituting equation $\left( iii \right)$ in $\left( iv \right)$ we get,
${{\left( \dfrac{2b}{k} \right)}^{2}}=\dfrac{a}{h}$
${{k}^{2}}=\dfrac{4{{b}^{2}}}{a}h$
This is the required locus.
Now since $\left( h,k \right)$ are general points on our locus we can replace $h$ by $x$ and $k$ by $y$.
${{y}^{2}}=\dfrac{4{{b}^{2}}}{a}x$
This is the required locus which represents another parabola.

Note: Students have to think carefully while deciding which is the variable before eliminating it. In this question students might eliminate $a$ which will give them the wrong answer. Also they may use their own different techniques to eliminate the variable from the equations. Also, if they feel it is redundant to use $\left( h,k \right)$ first and then replace it as $\left( x,y \right)$ they may use $\left( x,y \right)$ from the start as well.