The normal to the curve $x^2+2 x y-3 y^2=0$, at (1, 1) :

meets the curve again the fourth quadrant

The equation of the curve is

$x^2+2 x y-3 y^2=0$                ......(i)

Differentiating with respect to x, we get

$2 x+2 y+2 x \frac{d y}{d x}-6 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x+y}{3 y-x} \Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=1$

The equation of the normal at (1, 1) is

$y-1=-1(x-1)$ or, $x+y=2$                   ......(ii)

The x-coordinates of the points of intersection of (i) and (ii) are given by

$x^2+2 x(2-x)-3(2-x)^2=0$

$\Rightarrow 4 x^2-16 x+12=0$

$\Rightarrow x^2-4 x+3=0 \Rightarrow(x-3)(x-1)=0 \Rightarrow x=1,3$

Now, $x=1$ and $x+y=2 \Rightarrow y=1$

$x=3$ and $x+y=2 \Rightarrow y=-1$

Hence, the normal to the curve meets it again at $(3,-1)$ which lies in the fourth quadrant.