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# The locus of the foot of perpendicular drawn from the centre of the ellipse $x^2+ 3y^2=6$ on any tangent to it is:

$\begin{array}{1 1}(A)\;(x^2-y^2)^2=6x^2+2y^2 \\ (B)\;(x^2-y^2)^2=6x^2-2y^2 \\(C)\;(x^2+y^2)^2=6x^2+2y^2 \\(D)\;(x^2+y^2)^2=6x^2-2y^2 \end{array}$

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## 1 Answer

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The locus of the foot of perpendicular drawn from the centre of the ellipse $x^2+ 3y^2=6$ on any tangent to it is: $(x^2+y^2)^2=6x^2+2y^2$
Hence C is the correct answer.
answered Apr 7, 2014 by

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