In cartesian coordinates, the curve $ \Psi(t) $ is defined by: $$ \Psi (t)= \begin{pmatrix} 5sin(t)cos({\frac 35t})\\ 4cos(t)\\ 5sin(t)sin({\frac 35t})\\ \end{pmatrix} $$ $$ t \in [-\frac{\Pi}{2},\frac{\Pi}{2}] $$ Proof that the curve is always lying on an ellipsoid
$$ {\frac{(x^2+z^2)}{(A^2)}+\frac{(y^2)}{(B^2)}=1} $$
and determine A and B.
Let $x=5sin(t)cos(\frac{3}{5} t)$, $y=4cos(t)$, $z=5sin(t)sin(\dfrac{3}{5}t)$, then:
$$\frac{(x^2+z^2)}{A^2}+\frac{(y^2)}{B^2}=\frac{25sin^2(t)cos^2(\frac{3}{5} t) + 25sin^2(t)sin^2(\dfrac{3}{5}t)}{A^2}+\frac{16cos^2(t)}{B^2}$$ $$=\frac{25sin^2(t)}{A^2}+\frac{16cos^2(t)}{B^2}$$
If $A=5$ and $B=4$ then
$$\frac{(x^2+z^2)}{A^2}+\frac{(y^2)}{B^2}=\frac{25sin^2(t)}{25}+\frac{16cos^2(t)}{16}=1$$
