Find the equation of the normal to the curve at the point (1, -1 ): 10yx^2 + 6x - 2y + 3 = x^3

Firstly, an expression for dy/dx needs to be found to allow us to find the gradient of the normal. As a normal is a straight line, the equation y-y₁=m(x-x₁) can be used to find its equation. We are given x₁ and y₁ in the question and we want an equation in terms of x and y so we only need to find m, which is the gradient of the line. dy/dx gives the gradient of the tangent at a point, and the gradient of the normal at any point is the negative reciprocal of the gradient of the tangent, -dx/dy. Differentiating both sides of this equation with respect to x (using implicit differentiation) gives dy/dx = (3x^2 -20xy -6)/(10x^2 -2), leading to -dx/dy = (2-10x^2)/(3x^2-20xy-6). After substituting in the values of x and y given in the question, m=-dx/dy=-8/17. We can then plug this in the to equation we have for a straight line: y+1 = (-8/17)(x-1). It would be acceptable to leave the expression like this, however I would recommend rearranging to the form y=(-8/17)x-9/17.