Official solution
x2 + 2xy -- 3y2 = 0
Differentiate the curve
2x + 2y + 2xy' -- 6yy' = 0
\[ \Rightarrow \] \[ \Rightarrow \] x + y + xy' -- 3yy' = 0
\[ \Rightarrow \] y'(x -- 3y) = -- (x + y)
\[ \Rightarrow \] y' = \[{{x + y} \over {3y - x}}\]
Slope of normal = \[ - {{dx} \over {dy}}\] = \[{{x - 3y} \over {x + y}}\]
\[ \therefore \] \[{\left( { - {{dx} \over {dy}}} \right)_{\left( {2,2} \right)}}\] = \[{{2 - 6} \over {2 + 2}}\] = -1
Normal at (2, 2)
y -- 2 = -- 1 (x -- 2)
\[ \Rightarrow \] y + x = 4
\[ \therefore \] Perpendicular distance from (0,0)
= \[\left| {{{0 + 0 - 4} \over {\sqrt 2 }}} \right|\] = \[{2\sqrt 2 }\]