Parametrization for the curve on cylinder $y = 7 - x^4$ that passes through the point $(0, 7, -3) $when t = 0 and is parallel to the xy-plane

Question

Can you help me?

So far I have turned $y = 7-x^4$ into $\langle1, 1, 0\rangle$ and used it to make the equation $L = (0, 7, -3) + t(1, 1, 0)$. I know this is wrong, but I just don't know what, and I know it has to do with $y = 7-x^4$.

Thank you in advance.

Answer 1

When you choose your curve, $z$ can take any value, so just choose $f(t)=(t, 7-t^4,-3)$. Clearly, the curve is parallel to the xy-plane (a curve is parallel to a plane if it lies in a plane parallel to the plane)...no reparametrization is necessary and certainly we don't need to think about lines through the point $(0,7,-3)$.