$x^{2}+2\, x y-3 y^{2}=0$
diff. w.r.t. $x$
$2 x+2 x\left(y'\right)+2 y-6\, y\, y'=0$
$2+2\, y'+2-6 \,y'=0$
$4 y'=4$
$y'=1$
$\Rightarrow $slope of normal $=-1$
So equation becomes
$y-1=-1(x-1)$
$x+y=2$
solving it with curve
$x^{2}+2 x y-3 y^{2}=0$
$x^{2}+2 x(2-x)-3(2-x)^{2}=0$
$x^{2}+4 x-2 x^{2}-3\left(x^{2}-4 x+4\right)=0$
$-4 x^{2}+16 x-12=0$
$x^{2}-4 x+3=0$
$(x-1)(x-3)=0$
$x=1,3$
$\Rightarrow y=1,-1$
thus second point of intersection is $(3,-1)$ is in $4^{\text {th }}$ qud.