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Hint: Consider the equations of parabola and then write the equations of tangent and poles, after that try to compare its coefficients.

Complete step-by-step answer:

Let’s consider the equations of parabolas,

\[A\Rightarrow {{y}^{2}}=4ax\]

\[B\Rightarrow {{y}^{2}}=-4ax\]

They have the same vertex and axis but have their concavities turned in opposite directions.

So, now we consider tangent to parabola A,

The equation of tangent is:

$ty=x+a{{t}^{2}}...........(i)$

And, now we consider the equation of pole with respect to parabola ‘B’,

The equation of pole is:

$yK=-2ax-2ah........(ii)$

Now comparing the coefficients of the equation (i) and (ii), we get

$\dfrac{t}{K}=-\dfrac{1}{2a}=\dfrac{a{{t}^{2}}}{-2ah}......(iii)$

Now, consider part by part of the equation (iii)

$\dfrac{t}{K}=-\dfrac{1}{2a}$

$\Rightarrow t=-\dfrac{K}{2a}..........(iv)$

Now consider the last two parts of the equation (iii), we get

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

Cancelling the like terms and keeping ‘t’ on one side, we get

$\begin{align}

& \Rightarrow {{t}^{2}}=\dfrac{h}{a} \\

& \Rightarrow t=\sqrt{\dfrac{h}{a}}..........(v) \\

\end{align}$

Equating equation (iv) and (v), we get

$-\dfrac{K}{2a}=\sqrt{\dfrac{h}{a}}$

Taking square on both sides, we get

$\dfrac{{{K}^{2}}}{4{{a}^{2}}}=\dfrac{h}{a}$

Cancelling the like terms, we get

\[\Rightarrow {{K}^{2}}=4ah\]

This is the locus of the poles with respect to B of tangents to A.

Substituting ‘K’ by ‘y’ and ‘h’ by ‘x’, we get the locus as

${{y}^{2}}=4ax$

This is the same as the equation of the parabola A.

Therefore, the locus of poles with respect to B of tangents to A is the parabola A.

Hence proved

Note: In these types of questions students have problems regarding the equations of tangents and poles of parabola. Students have to know these equations by heart to solve the problems easily.

Another mistake is once the student gets the locus equation, i.e., \[{{K}^{2}}=4ah\], they forget to replace ‘K’ by ‘y’ and ‘h’ by ‘x’, and gets stuck in the final part.

Complete step-by-step answer:

Let’s consider the equations of parabolas,

\[A\Rightarrow {{y}^{2}}=4ax\]

\[B\Rightarrow {{y}^{2}}=-4ax\]

They have the same vertex and axis but have their concavities turned in opposite directions.

So, now we consider tangent to parabola A,

The equation of tangent is:

$ty=x+a{{t}^{2}}...........(i)$

And, now we consider the equation of pole with respect to parabola ‘B’,

The equation of pole is:

$yK=-2ax-2ah........(ii)$

Now comparing the coefficients of the equation (i) and (ii), we get

$\dfrac{t}{K}=-\dfrac{1}{2a}=\dfrac{a{{t}^{2}}}{-2ah}......(iii)$

Now, consider part by part of the equation (iii)

$\dfrac{t}{K}=-\dfrac{1}{2a}$

$\Rightarrow t=-\dfrac{K}{2a}..........(iv)$

Now consider the last two parts of the equation (iii), we get

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

Cancelling the like terms and keeping ‘t’ on one side, we get

$\begin{align}

& \Rightarrow {{t}^{2}}=\dfrac{h}{a} \\

& \Rightarrow t=\sqrt{\dfrac{h}{a}}..........(v) \\

\end{align}$

Equating equation (iv) and (v), we get

$-\dfrac{K}{2a}=\sqrt{\dfrac{h}{a}}$

Taking square on both sides, we get

$\dfrac{{{K}^{2}}}{4{{a}^{2}}}=\dfrac{h}{a}$

Cancelling the like terms, we get

\[\Rightarrow {{K}^{2}}=4ah\]

This is the locus of the poles with respect to B of tangents to A.

Substituting ‘K’ by ‘y’ and ‘h’ by ‘x’, we get the locus as

${{y}^{2}}=4ax$

This is the same as the equation of the parabola A.

Therefore, the locus of poles with respect to B of tangents to A is the parabola A.

Hence proved

Note: In these types of questions students have problems regarding the equations of tangents and poles of parabola. Students have to know these equations by heart to solve the problems easily.

Another mistake is once the student gets the locus equation, i.e., \[{{K}^{2}}=4ah\], they forget to replace ‘K’ by ‘y’ and ‘h’ by ‘x’, and gets stuck in the final part.