Osculating plane Consider the curve: $$x=p \sqrt{p^2-q^2}\cos t; \ \ \ y=q\sqrt{p^2-q^2}\ \ (1+\sin t) \ \ \ z=(p^2-q^2)(1+\sin t) \ while \ p>q$$
Check that the osculating plane to the curve is one of the planes of circular section in the paraboloid: $$\frac{x^2}{p^2}+\frac{y^2}{q^2}=2z $$
I have tried to solve this question by first finding the osculating plane but I couldnt because the point on which I am supposed to find the osculating plane is not given. So, do I have to take an arbitrary point on the curve ? Also I did compute the unit normal, unit tangent and binormal vectors, which are $$ T(t)=(-\sin t,\frac {q}{p}\cos t,\frac{\sqrt{p^2-q^2}}{p}\cos t) $$ $$N(t)=(-\cos t,-\frac{q}{p}\sin t,\frac{\sqrt{p^2-q^2}}{p}\sin t)$$ $$ B(t)=(0,-\frac{\sqrt{p^2-q^2}}{p},\frac{q}{p}\ )$$
Can someone please tell me what should I do next ?
 A: $B(t)$ is constant, so, the curve is a flat one and lies in the plane orthogonal to $B$,$O:-\frac{\sqrt{p^2-q^2}}{p}Y+\frac{q}{p}Z=0 $. The osculating plane is the same the curve is, so is, for any $t$ the osculating plane is $O$. We have to prove that the intersection of $O$ with $P$ is a circle.
$\begin{cases}
 -\sqrt{p^2-q^2}y+qz=0\\
  \dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=2z
\end{cases}$
Let $ y=q\sqrt{p^2-q^2}(1+\sin t)$, then 
$z=\dfrac{\sqrt{p^2-q^2}}{q}y=(p^2-q^2)(1+\sin t) $
$\dfrac{x^2}{p^2}=2z-\dfrac{y^2}{q^2}=2(p^2-q^2)(1+\sin t)-(p^2-q^2)(1+\sin t)^2=$
$=(p^2-q^2)(2+2\sin t-1-2\sin t-\sin^2t)$
$x=p\sqrt{p^2-q^2}\cos t$
It's the very same curve!
$\begin{cases}
 x=p \sqrt{p^2-q^2}\cos t\\
 y=q\sqrt{p^2-q^2}(1+\sin t)\\
 z=(p^2-q^2)(1+\sin t)
\end{cases}$
A plane curve is a circle if it's curvature $k$ is constant. In general $T'=KN$
$T(t)=(-\sin t,\dfrac {q}{p}\cos t,\dfrac{\sqrt{p^2-q^2}}{p}\cos t)$
$T'(t)=(-\cos t,-\dfrac {q}{p}\sin t,-\dfrac{\sqrt{p^2-q^2}}{p}\sin t)$
$T'(t)=N(t)$ so is, $K=1$.
Added
We are not working with arclength parametrised curves, so $K$ is not the curvature. This is $k=\dfrac{1}{p\sqrt{p^2-q^2}}$. $T'=KN$ with $K$ constant works because the "speed" of the curve is constant too.
