The question:

Show that the curve $\displaystyle r(t) = t^2\underline{i} + t\underline{j} + (5t - 4)\underline{k}$ is normal to the surface $\displaystyle 2x^2 + y^2 + 5z^2 = 8$ at the point (1, 1, 1).

My attempt:

I found the tangent plane for the second surface, which is:

$\displaystyle 2x + y + 5z = 8$

I then noticed that the curve is at point (1, 1, 1) when t = 1. I differentiated the curve, substituted t = 1, and got:

$\displaystyle 2\underline{i} + \underline{j} + 5\underline{k}$

Which is the same as the tangent plane for the second surface. I'm confused, how can it be normal to the surface if it has the same tangent? Or have I calculated it wrong?