The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2 x y-3 y^2=0$ at the point $(2,2)$ is

The perpendicular from the origin to the normal of the curve $x^2 + 2xy - 3y^2 = 0$ at $(2,2)$ is determined.

Implicit differentiation gives the slope of the tangent at any point:
$\frac{dy}{dx} = \frac{x + y}{3y - x}$

At $(2,2)$, the slope of the tangent is $m_t = \frac{2 + 2}{3(2) - 2} = 1$, so the slope of the normal is $m_n = -1$.

The equation of the normal is then $y - 2 = -1(x - 2)$ or $x + y - 4 = 0$.

The perpendicular distance from the origin to this line is
$\frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

The length is $2\sqrt{2}$ units, corresponding to option C.