The point on the curve where y=1, we will have x=2, by substituting y=1 in the given equation and solving for x. Hence we have to write the equation of tangent at the point (2,1) of the curve -2y³ +5xy= -2x+12.

Differentiating the equation with respect to x, we will have -6y² +5x +5x=-2.

Substituting x=2, y=1 , we will have = 7 (Slope of the tangent at point 2,1)

Hence the equation of the Tangent will be y-1= 7(x-2), Or y=7x-13