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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 :

$(a)\;(x^{2}+y^{2})^{2}=6x^{2}+2y^{2} \qquad(b)\;(x^{2}+y^{2})^{2}=6x^{2}-2y^{2} \qquad(c)\;(x^{2}-y^{2})^{2}=6x^{2}+2y^{2} \qquad(d)\;(x^{2}-y^{2})^{2}=6x^{2}-2y^{2}$

Can you answer this question?