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The locus of the foot of perpendicular d...

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 (1) `(x^2-y^2)^2=""6x^2+""2y^2` (2) `(x^2-y^2)^2=""6x^2-2y^2` (3) `(x^2+y^2)^2=""6x^2+""2y^2` (4) `(x^2+y^2)^2=""6x^2-2y^2`

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

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The correct Answer is:
C
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