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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

If the polar of $y^2=4ax$ is always touching the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ then locus of the pole is :

$\begin{array}{1 1}(A)\;4a^2x^2-b^2y^2=4a^4 \\(B)\;a^2x^2-b^2y^2=a^4 \\(C)\;4a^2x^2+b^2y^2=4a^4 \\(D)\;none\;of\;these \end{array}$

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1 Answer

Let coordinates of the pole be $ (h,k)$ then equation of the polar of $y^2=4ax$ is $ky=2a(x+h)$
$y=\large\frac{2a}{k}$$x+\large\frac{2ah}{K}$
Since line (1) is touching the ellipse
$\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
$\large\frac{4a^2b^2}{k^2}=a^2.\large\frac{4a^2}{K^2}$$+b^2$
Required locus is $4a^2x^2=4a^4+b^2y^2$
=> $ 4a^2x^2-b^2y^2=4a^4$
Hence A is the correct answer.
answered Apr 10, 2014 by meena.p
 

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