A curve has parametric equations x=t(t-1), y=4t/(1-t). The point S on the curve has parameter t=-1. Show that the tangent to the curve at S has equation x+3y+4=0.

To anwser this question we need to find a linear equation of the form y=mx+c which we can rearrange to give the desired equation. Firstly, we must find the gradient at the point S given by dy/dx. Using the parametric equations we have dx/dt=2t-1 by standard differentiation and dy/dt=4/(1-t)^2 by the quotient rule of differentiation. We can know find the gradient of the line at the point S. dy/dx=(dy/dt)/(dx/dt)=4/(2t-1)(1-t)^2 and t=-1 at the point S leaving us with gradient m=-1/3. We now have an equation of the form y=-(1/3)x+c. We know x=2, y=-2 at S given t=-1. We can now find c by substituting the values of x and y into our equation and rearranging to find c=-4/3. We are left with y=-(1/3)x-(4/3) which if we multiply through by 3 and rearrange to get a 0 on one side of the equation we are left with x+3y+4=0 which is what we want.