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Normal at (2, 2) to curve x^2 + 2xy – 3y...

Normal at `(2, 2)` to curve `x^2 + 2xy – 3y^2 = 0` is L. Then perpendicular distance from origin to line L is

A

`2sqrt2`

B

2

C

`4sqrt2`

D

`sqrt2`

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The correct Answer is:
A
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