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The locus of the foot of the perpendicul...

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

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)`

Text Solution

Verified by Experts

Let the foot of perpendicular be P(h,k)
Equation of tangetn with slope m passing through P(h,k) is `y=mx+-sqrt(6m^(2)+2), "where" m=(h)/(k)`
`rArr sqrt((6h^(2))/(k^(2))+2)=+-(h^(2)+k^(2))/(k)rArr6h^(2)+2k^(2)=(h^(2)+k^(2))^(2)`
So, requried locus is `6x^(2)+2y^(2)=(x^(2)+y^(2))^(2)`
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