A curve is defined by the parametric equations x=t^2/2 +1 and y=4/t -1. Find the gradient of the curve at t=2 and an equation for the curve in terms of just x and y.

To find the gradient of the curve at t=2 we need to find an expression for dy/dx and then substitute in for t=2. We can make use of the chain rule to find this expression because dy/dx = (dy/dt)/(dx/dt) and these derivates are easier to calculate. From the parametric equations, dx/dt = t and dy/dt = -4/t^2. Therefore dy/dx = (-4/t^2)/t = -4/t^3. Now t=2 can be substituted in to find that the gradient at this point is -1/2.

In order to find an equation for the curve in terms of just x and y, we need to eliminate the t variable. Rearranging the y equation tells us that t=4/(y+1). Now this can just be substituted into the x equation to give x=8/(y+1)^2 + 1.