How to show that the curve $ (x,y,z) = \langle \cos t, \sin t, c\sin t\rangle $ is an ellipse?

Question

Show that the curve $$(x,y,z) = \langle \cos t, \sin t, c\sin t\rangle $$ is an ellipse in the plane it lies on.

$$x^2 + y^2 = (\sin t)^2 + (\cos t)^2 = 1$$ $$x^2 + (z/c)^2 = (\sin t)^2 + (\cos t)^2 = 1$$ $$\implies x^2 + y^2 = x^2 + (z/c)^2$$ $$\implies y^2 - (z/c)^2 = 0$$ $$x^2 + y^2 + 0 = 1$$ $$\implies x^2 + y^2 + y^2 - (z/c)^2 = 1$$ $$\implies x^2 + 2y^2 - (z/c)^2 = 1$$

This however is an equation for an one sheeted hyperboloid, not a curve. Where did I mess up?

Answer 1

What you've written is true, however you're forgetting that $z/c=y$. It's kind of like saying $x^2+y^2+(z/c)^2=1$ is an ellipsoid but $x^2+2y^2-(z/c)^2=1$ is a hyperbeloid.

Now, you have:

$x^2+(z/c)^2=1$ and $x^2+y^2=1$. Now add the two together and you get your "ellipsoid", but your system is parametrized by $t$ only so it's a curve in 3D which lies on an ellipsoid.

Answer 2

Here's another way. Define the following variables: $$ \begin{align} \tilde x&=x\\ \tilde y&=y \cos\alpha -z \sin\alpha\\ \tilde z&=-y\sin\alpha +z \cos\alpha \end{align}$$ with $\alpha=\tan^{-1} (c)$. With these variables, the curve is of the form $$(\tilde x, \tilde y, \tilde z)= \left(\cos t,\frac{\sin t}{\cos\alpha},0\right)= \left(\cos t,\frac{\sin t}{\sqrt{1+c^2}},0\right)$$