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If polar of the parabola $x^2=4ay$ is always touching the parabola $y^2=4ax$ then locus of the pole is :

$\begin{array}{1 1}(A)\;x-y=2a^2 \\(B)\;xy=-2a^2 \\(C)\;xy=a^2 \\(D)\;none\;of\;these \end{array}$

1 Answer

Let coordinates of the pole be $(h,k)$ then equation of the polar of $x^2=4ay$
$hx= 2a(y+k)$
=> $ y=\large\frac{h}{2a} $$(x-k)$
Since polar (1) is touching the parabola $y^2=4ax$
$-K= \large\frac{a}{h/2a}$
=> $hk= -2a^2$
Required locus is
$xy =-2a^2$
Hence B is the correct answer.
answered Apr 9, 2014 by meena.p

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