Need help showing important tangent plane property:

[ozone]

Homework Statement

Consider the tangent surface of some regular differentiable curve given as $X(t,v) = \alpha(t) + v \alpha'(t)$. Show that the tangent planes along X(t,constant) are equal.

Homework Equations

$N = \frac{X_{t} \wedge X_{v}}{|X_{t} \wedge X_{v}|}$

The general tangent plane equation, $T_{t_0}(S) = N|_{t_0, v_0} \cdot( (t - t_0) - (v-v_0)) = 0$

The Attempt at a Solution

Using the equation above I solved for $N = \hat{v} b(s)$ where b(s) is the binormal to the curve.

However I wasn't sure if this was correct since it does not seem like I am close to answering the problem. I wanted to reach the conclusion that $\frac{dN}{dt} = 0$ along this curve, but this seemed to be untrue. Another idea was to show that the tangent plane equation is the same for all initial points $t_0$ but this too seemed incorrect.

I am really stuck at this point and would love it if anyone could point me in the right direction

 

[ozone]

Edit: I realized v is just a constant so what we really should have is something like $N = \pm b(t)$. Also I forgot to mention I had a factor of $|\alpha(t)|^4$ so that $N = \pm |\alpha(t)|^4 b(t)$ but I assumed that the curve is parameterized by arc-length so that I could ignore this term.. I'm not sure that it should make any difference by including it.

Also I should note that I wrote the tangent plane equation down very wrong, it should be given by

$T_{t_0,v_0}(S) = N|_{t_0,v_0} \cdot ((x,y,z) - X(t_0,v_0))$

Which would simplify to

$T_{t_0,v_0}(S) = \pm |\alpha(t_0)|^4 (b(t_0) \cdot (x,y,z) - b(t_0) \cdot \alpha(t_0) )$

Which we would then want to show is not changing as $t_0$ varies