Given the paraboloid z = 6 - x - x2 -2y2 and the plane x = 1, find curve of intersection and the parametric equations of the tangent line to this curve at point (1,2,-4). So I plugged x=1 into the paraboloid equation and got z = 4-2y2. Then I take the derivative of the curve of intersection: dz/dy = -4y Then I plug in 2 for t since y=t=2 at the given point and get the slope -8. The parametric equation of a line is r = r0 +tv I have r0 = (1,2,-4) and I have a slope -8. I'm not sure how to get a vector.