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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/a^2+y^2/b^2=1` on any tangent is given by `(x^2 + y^2)^2 = lx^2+my^2`, where:

A

`l=a^2, m=b^2`

B

`l=-a^2, m=b^2`

C

`l=-a^2, m=-b^2`

D

`l=a^2, m=-b^2`.

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