Normal at `(2, 2)` to curve `x^2 + 2xy – 3y^2 = 0` is L. Then perpendicular distance from origin to line L is

A normal is drawn to (x)/(9)+(y)/(4)=2 .at a point (3,2) .The perpendicular distance of normal from origin is :

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

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

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

Find the equation of the normal to the curve y=e^(2x)+x^(2)" at "x=0 . Also find the distance from origin to the line.

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

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

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