JEE Main · 2015 · 04 Apr 2015 · Offline · Application of Derivatives

The 'curve' actually factorises: $x^2+2xy-3y^2=(x-y)(x+3y)=0$, a pair of straight lines. Implicit differentiation, $2x+2y+2x{y}^{\prime }-6y{y}^{\prime }=0$, gives slope $y^{\prime }=1$ at $(1,1)$, so the normal there has slope $-1$ and equation $x+y=2$. This normal already meets the branch $x-y=0$ at $(1,1)$, so look to the other branch $x+3y=0$: solving it with $x+y=2$ gives $y=-1$, $x=3$. The point $(3,-1)$ lies in the fourth quadrant.