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

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

Find the slopes of the tangent and the normal to the curve x^(2)+3y_(y)^(2)=5 at (1,1)

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 equation of normal to the curve 2x^(2)+3y^(2)-5=0 at (1,1) is

Find the equations of tangent and normal to the curve 2x^(2)-xy+3y^(2)=18at(3,1)

Find the slopes of the tangent and the normal to the curve x^2+3y+y^2=5 at (1,\ 1)

Find the slopes of the tangent and the normal to the curve x^2+3y+y^2=5 at (1, 1)

Find the slopes of the tangent and the normal to the curve x^2+3y+y^2=5 at (1,\ 1)

Find the slopes of the tangent and the normal to the curve x^2+3y+y^2=5 at (1, 1)

The equation of the normal to the curve 2x^2-xy+3y^2=18 at (3,1) is