Prove that the locus of the poles of tangents to the parabola $y^2=4ax$ with respect to the circle $x^2+y^2-2ax=0$ is the circle $x^2+y^2-ax=0$.
I have encountered this question from SL Loney.I have its solution in front of me,but i could not understand.
Let $(h,k)$ be the coordinates of the pole,then the polar of $(h,k)$ with respect to the circle
$$x^2+y^2-2ax=0$$
is
$$xh+yk-a(x+h)=0$$
or
$$y=x\left(-\frac{h}{k}+\frac{a}{k}\right)+\frac{ah}{k}\tag{1}$$
The equation of any tangent on the given parabola is
$$y=mx+\frac{a}{m}\tag{2}$$
Equations $(1)$ and $(2)$ represent the same straight line,hence equating the coefficients,
$$m=\left(-\frac{h}{k}+\frac{a}{k}\right) \;\; \text{and} \;\; \frac{a}{m}=\frac{ah}{k}$$
Eliminate $m$ between the above two equations and locus of $(h,k)$ is $h^2+k^2=ah$.
Generalizing, we get the locus as $x^2+y^2=ax$.
My question is why equations $(1)$ and $(2)$ represent the same line? Equation $(1)$ is the polar of $(h,k)$ with respect to the circle $x^2+y^2-2ax=0$ and equation $(2)$ is tangent to the parabola.
Best Answer
In the following part of the question,
note that the $\left(\quad\right)$ part is the polar (with respect to the circle $x^2+y^2-2ax=0$). (in other words, the question means that the tangents to the parabola $y^2=4ax$ is the polar with respect to the circle $x^2+y^2-2ax=0$.) (take a look at polar in mathworld)
$(1)$ is the polar of the pole $(h,k)$ with respect to the circle $x^2+y^2−2ax=0$.
$(2)$ is the tangents to the parabola $y^2=4ax$.
So, from the question, we (can/have to) have $(1)=(2)$.