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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^(20) +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:
B
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