A) \[2\sqrt{2}\]
B) \[\sqrt{2}\]
C) \[4\sqrt{2}\]
D) \[2\]
Correct Answer: A
Solution :
Given curve \[{{x}^{2}}+3xy-xy-3{{y}^{2}}=0\] \[\Rightarrow \]\[\left( x-y \right)\left( x+3y \right)=0\] \[\Rightarrow y=x\text{ }and\text{ }y=-\frac{x}{3}\] \[\therefore \] Normal pass through (2, 2) and is perpendicular to line \[x-y=0\] Let normal is \[x+y+\lambda =0\] \[\Rightarrow \text{ }\left| X=-4 \right|\] \[\therefore \] Perpendicular distance = \[\left| \frac{-4}{\sqrt{2}} \right|=2\sqrt{2}\]
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