Polar of a point locus of the point? I'm trying to solve this problem but can't understand what is meant by this "polar"
The question is as follows.,
"If the polar of any point with respect to the parabola $y^2=4ax$ touches the circle $y^2+x^2=4a^2$, show that the locus of the point is the curve $x^2-y^2=4a^2$
Thank you. 
 A: The term "polar" is standard in the study of conic sections, so it is probably defined in your class notes or textbook. Anyway, the polar of a point $P$ with respect to a conic $C$ is the line that passes through the two points where tangents from $P$ meet $C$.
There is a (pretty poor) description here.
I'll look for a better one.
This page is better, though it only deals with pole/polar of a circle.
A: Assume the point to be $P(h,k)$ whose polar w.r.t. parabola $y^2 = 4ax$ is tangent to circle $x^2 +y^2 = 4a^2$, say at $Q(\alpha,\beta)$. Then writing $T = 0$ (equation of polar) for $P$ w.r.t. parabola $y^2 = 4ax$, we get 
$yk = 4a\left(\frac{x + h}{2}\right)\Rightarrow yk = 2a(x + h)\Rightarrow 2ax-ky+2ah = 0\;...(1)$
Now equation of tangent to circle $x^2 +y^2 = 4a^2$ at $Q$ is $$ \alpha x + \beta y -4a^2 = 0\; ...(2) $$ 
Since $(1)$ and $(2)$ represent equation of same straight line, hence by comparison,
$ \frac{\alpha}{2a} = \frac{\beta}{-k} = \frac{-4a^2}{2ah} $ $\Rightarrow \alpha = {-4a^2 \over h} ;\beta = {2ak \over h}$
Also $\alpha^2 + \beta^2 = 4a^2 \Rightarrow \left({-4a^2 \over h} \right)^2 + \left({2ak \over h }\right)^2 = 4a^2 \Rightarrow 4a^2 + k^2 = h^2 \Rightarrow h^2 - k^2 =4a^2 $ 
Thus the locus of $P(h,k)$ is $x^2 -y^2 = 4a^2$.
