Question

Find the tangent line to the parametric curve (t^3+2t,t^2+1) at the point where t = -1, and then give the value of x where this line intersects the line given by (-1+2t,3-t).

Answer 1

dy/dx=(dy/dt)/(dx/dt)=2t/(3t^2+2). When t=-1, dy/dx=-2/5. The tangent line is given by y=-2x/5+b where b is found by plugging in (x,y) when t=-1: x=t^3+2t=-3; y=t^2+1=2 so 2=6/5+b and b=4/5, so 5y=4-2x is the tangent line.

The parametric gives x=-1+2t, y=3-t so t=3-y and x=-1+6-2y=5-2y or 2y+x=5. The tangent line is 2x+5y=4. 4y+2x=10=4y+4-5y=4-y, so y=-6. This makes x=5-2y=5+12=17.