The locus of the foot of perpendicular drawn from the center of the ellipse x^{2} + 3y^{2} = 6 on any tangent to it is

- (x
^{2}- y^{2})^{2}= 6x^{2}+ 2y^{2} - (x
^{2}- y^{2})^{2}= 6x^{2}- 2y^{2} - (x
^{2}+ y^{2})^{2}= 6x^{2}- 2y^{2} - (x
^{2}+ y^{2})^{2}= 6x^{2}+ 2y^{2}

**Answer**

Let the foot of perpendicular be P(h,k)

Equation of tangent with slope m passing P(h, k) is

y = mx ± √(6m^{2} + 2), where m = -h/k

√(6h^{2}/k^{2} + 2) = (h^{2} + k^{2})/k

6h^{2} + 2k^{2} = (h^{2} + k^{2})^{2}

**The correct option is D.**