2

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.

RinW
  • 335
  • 6
  • 18

2 Answers2

1

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.

bubba
  • 43,483
  • 3
  • 61
  • 122
0

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 $x^2 -y^2 = 4a^2$.

Arpan Mathur
  • 1