# 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:

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

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.