Prove that the points of a circular helix such that the osculator plane in them pass through a point $p$, are all in the same plane.
How can I prove this statement? I've found the osculator plane in any point of a circular helix, but I do not know how to use it.
After some appropriate rotation and translation, a circular helix can be expressed in parametric equations:
$x(t)=a\sin t$
$y(t)=a\cos t$
$z(t)=bt$
for some $a,b\in\mathbb{R}$. We compact this by writing $r(t)=x(t)\mathbf{\hat{i}}+y(t)\mathbf{\hat{j}}+z(t)\mathbf{\hat{k}}$.
We can now find the equations of the tangent vector $\mathbf{T}(t)$ and normal vector $\mathbf{N}(t)$.
$\mathbf{T}(t)=\dfrac{r'(t)}{||r'(t)||}=\dfrac{a\cos t\mathbf{\hat{i}}-a\sin t\mathbf{\hat{j}}+b\mathbf{\hat{k}}}{\sqrt{a^2\cos^2 t+a^2\sin^2t+b^2}}=\dfrac{1}{\sqrt{a^2+b^2}}(a\cos t\mathbf{\hat{i}}-a\sin t\mathbf{\hat{j}}+b\mathbf{\hat{k}})$
$\begin{align} \mathbf{N}(t)=\dfrac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||}=\dfrac{\dfrac{1}{\sqrt{a^2+b^2}}(-a\sin t\mathbf{\hat{i}}-a\cos t\mathbf{\hat{j}})}{\dfrac{1}{\sqrt{a^2+b^2}}\sqrt{a^2\sin^2 t+a^2\cos^2 t}}&=\dfrac{1}{a}(-a\sin \mathbf{\hat{i}}-a\cos t\mathbf{\hat{j}})\\&=-\sin t\mathbf{\hat{i}}-\cos t\mathbf{\hat{j}}\end{align}$
We can now calculate a normal to the osculating plane = $k(\mathbf{T}(t)\times \mathbf{N}(t))$, for some $k\in\mathbb{R}$. For the sake of ease of computation, we will choose the normal that is
$\begin{align}(a\cos t\mathbf{\hat{i}}-a\sin t\mathbf{\hat{j}}+b\mathbf{\hat{k}})\times(-\sin t\mathbf{\hat{i}}-\cos t\mathbf{\hat{j}})&= \left|\begin{matrix} i & j & k\\ a\cos t & -a\sin t &b\\ -\sin t & - \cos t&0\\ \end{matrix}\right| \\&=b\cos t\mathbf{\hat{i}}-b\sin t\mathbf{\hat{j}}-a\mathbf{\hat{k}} \end{align}$
Hence the equation for the osculating plane at point $\mathbf{P}(a\sin t,a\cos t,bt)$ is $$b\cos t(x-a\sin t)-b\sin t(y-a\cos t)-a(z-bt)=0$$ which can be simplified to $$(b\cos t)x-(b\sin t)y+abt=az$$
Finally, we have that the plane passes through point $p=(x_1,y_1,z_1)$ for all $t$. Hence $$(b\cos t)x_1-(b\sin t)y_1+abt=az_1$$ for all $t$. Note that the right hand side of this equation is constant, while the left hand side can be written as $b((\cos t)x_1-(\sin t)y_1+at)$. For the moment assuming that $a\ne0$ gives us that $(\cos t)x_1-(\sin t)y_1+at$ is never constant, so in order for the left hand side to be constant, we need $b=0$. Finishing up, if $a=0$, we have $(b\cos t)x_1=(b\sin t)y_1$ for all $t$, so $b=0$ unless $x_1=y_1=0$.
To conclude, we have that $b=0$ unless $a=x_1=y_2=0$. However, if $a=0$, we have that $\mathbf{N}(t)$ doesn't exists, so there is no osculating pane. This leaves us with the conclusion that $b=0$, but when we have $b=0$, our original parametric equations become
$x(t)=a\sin t$
$y(t)=a\cos t$
$z(t)=0$,
so all the points of the circular helix are in the same plane.