A) meets the curve again in the third quadrant.
B) meets the curve again in the fourth quadrant.
C) does not meet the curve again.
D) meets the curve again in the second quadrant.
Correct Answer: B
Solution :
\[\frac{dy}{dx}=1\] \[-\frac{dy}{dx}\left( x,y \right)=-1\]equation of normal at (1,1) is \[\left( y-1 \right)=-\left( x-1 \right)\Rightarrow x+y=2\] Point of intersection of normal with the curve \[{{x}^{2}}+2xy-3{{y}^{2}}=0\] ?(1) \[x+y=2\] Elliminate y in both the equations \[x=1,3\]hence, \[\text{y}=\text{1},\text{1}\]\[(1,1)\And (3,-1)\] Hence, meets the curve again in the fourth quadrant.
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