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

The length of the perpendicular from the origin, on the normal to the curve, x^2 + 3xy - 10 y^2 = 0 at the point (2,1) 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 length of the perpendicular from the origin to the line 4x+3y-2=0

If length of the perpendicular from the origin upon the tangent drawn to the curve x^2 - xy + y^2 + alpha (x-2)=4 at (2, 2) is equal to 2 then alpha equals

The equation of the normal to the curve y=x(2-x) at the point (2, 0) is

The equation of normal to the curve, x^(2//3)+y^(2//3)=a^(2//3) at the point (a,0) is