The length of the perpendicular from the origin,

Given equation of curve is

$x^{2}+2 x y-3 y^{2}=0$

$\Rightarrow 2 x+2 y+2 x y^{\prime}-6 y y^{\prime}=0$

$\Rightarrow x+y+x y^{\prime}-3 y y^{\prime}=0$

$\Rightarrow y^{\prime}(x-3 y)=-(x+y)$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{3 y-x}$

Slope of normal $=\frac{-d x}{d y}=\frac{x-3 y}{x-3 y}$

Normal at point $(2,2)=\frac{2-6}{2+2}=-1$

Equation of normal to curve $=y-2=-1(x-2)$

$\Rightarrow x+y=4$

$\therefore \quad$ Perpendicular distance from origin

$=\left|\frac{0+0-4}{\sqrt{2}}\right|=2 \sqrt{2}$