the locus of the foot of perpendicular drawn from the centre of the ellipse `x^2+3y^2=6` on any tangent is

Options

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)`

Answer

A

Explanation

Let P(h,k) be the foot of perpendicular from the centre O (0,0) on any tangent `y=mx+sqrt(6m^(2)+2)`. Then,
`k=mh+sqrt(6m^(2)+2)and (k)/(h)xxm=-1`
` rArr (k-mh)^(2)=6m^(2)+2andm=-(h)/(k)`
`rArr (k+(h^(2))/(k))^(2)=(6h^(2))/(k^(2))+2" "["On eliminataing m"]`
`rArr (h^(2)+k^(2))=6h^(2)+2k^(2)`

Hence, the locus of (h, k) is `(x^(2)+y^(2))^(2)=6x^(2)+2y^(2)`.