Find the equation of the normal to the curve x^3 + 2(x^2)y = y^3 + 15 at the point (2, 1)

The first thing when seeing such a question is to find out what the question wants you to do and plan your method to do so.In this case, the question asks you to find the normal of a curve at a given point (2,1). And we know the normal lines are linked to tangents such that their gradient's product is -1. And tangent line is linked to differentiation. Since a point is given then all we need to do is to differentiate the given curve and sub in the value for the point to find the gradient of the tangent at point(2,1). But obviously it's impossible to rearrange the curve equation such that y and x are on different sides of the equation, therefore we would implicitly differentiate the entire equation by d/dx first and the nrearrange to find dy/dx. We different with respect to dx because this is the definition of the gradient of the tangent and would facilitate our further processes.And we would get: 3x^2 + (4xy+ 2x^2*(dy/dx))=3y^2*(dy/dx)in which we have applied the chain ruleThen we rearrange to get (3y^2-2x^2)*(dy/dx)=3x^2+4xytherefore dy/dx=(3x^2+4xy)/(3y^2-2x^2)sub in x=2, y=1dy/dx= (12+8)/(3-8)=20/(-5)=-4this is the gradient of the tangent, therefore the gradient of the normal is (-1)/(-4)=1/4and we again sub in x=2, y=1 into y=mx+cto find that 1=(1/4)*2+ctherefore c=1/2therefore the equation is y=x/4+(1/2) or 4y-x-2=0