I'm given the following curve that corresponds to a generatrix curve of a rotation surface.
$$ \gamma=\frac{x^2}{a^2}+\frac{z^2}{c^2}=1, y=0 $$
I'm asked the following among many questions: if this surface rotates around the X axis, ¿what surface and what equation do you get?
I'm having a bit of a trouble because I'm not given the values of $a$ and $c$. Or perhaps I'm misunderstanding the problem. I mean, if $a=c$ then I suppose I would get a circle, otherwise depending on the values it could be a horizontal or vertical ellipse, am I right? Or am I misunderstanding the problem? If so, what are the common steps to tackle this type of problem?
General case
Given a curve $\gamma : \left[u_i,\,u_f\right] \to \mathbb{R}^3$ of law:
$$ \gamma(u) := \left(x(u),\,y(u),\,z(u)\right) $$
and an axis of parametric equations:
$$ (x,\,y,\,z) = \left(x_0 + l\,t, \; y_0 + m\,t, \; z_0 + n\,t\right) \; \; \; \text{with} \; t \in \mathbb{R}\,, $$
where is $l \ne 0$.
Turning the curve around the axis of $\overline{v} \in [0,\,2\pi]$ generates a surface $\sigma : \left[u_i,\,u_f\right] \times \left[0,\,\overline{v}\right] \to \mathbb{R}^3$ of law:
$$ \sigma(u,\,v) := \left(\overline{x}(u,\,v),\,\overline{y}(u,\,v),\,\overline{z}(u,\,v)\right) $$
where:
$$ \begin{aligned} & \overline{x}(u,\,v) := x(u) + \frac{m\left(y(u) - \overline{y}(u,\,v)\right) + n\left(z(u) - \overline{z}(u,\,v)\right)}{l} \;; \\ & \overline{y}(u,\,v) := y_0 + m\,t(u) + r(u)\left(\frac{l}{\sqrt{l^2 + m^2}}\,\cos v - \frac{m\,n}{\sqrt{\left(l^2 + m^2\right)\left(l^2 + m^2 + n^2\right)}}\,\sin v\right) \;;\\ & \overline{z}(u,\,v) := z_0 + n\,t(u) + r(u)\,\sqrt{\frac{l^2 + m^2}{l^2 + m^2 + n^2}}\,\sin v \;; \\ \end{aligned} $$
with:
$$ \begin{aligned} & t(u) := \frac{l\left(x(u) - x_0\right) + m\left(y(u) - y_0\right) + n\left(z(u) - z_0\right)}{l^2 + m^2 + n^2} \;; \\ & r(u) := \sqrt{\left(x_0 + l\,t(u) - x(u)\right)^2 + \left(y_0 + m\,t(u) - y(u)\right)^2 + \left(z_0 + n\,t(u) - z(u)\right)^2} \;. \end{aligned} $$
Equations completely similar if we suppose $m \ne 0$ or $n \ne 0$.
Special case
If the axis coincides with the x-axis, obtainable by placing $(x_0,\,y_0,\,z_0) = (0,\,0,\,0)$ and $(l,\,m,\,n) = (1,\,0,\,0)$, we have:
$$ \sigma(u,\,v) = \left(x(u),\,\sqrt{y^2(u) + z^2(u)}\,\cos v,\,\sqrt{y^2(u) + z^2(u)}\,\sin v\right) $$
where, for a half rotation, $v \in [0,\,\pi]$.
Furthermore, by parameterizing the assigned curve in a natural way by placing $x(u) = a\,\cos u$, $y(u) = 0$, $z(u) = c\,\sin u$, with fixed $a,\,c > 0$ and $u \in [0,\,2\pi]$, we have:
$$ \sigma(u,\,v) = \left(a\,\cos u,\,c\,\sin u\,\cos v,\,c\,\sin u\,\sin v\right) $$
with $(u,\,v) \in [0,\,2\pi] \times [0,\,\pi]$.
This done, noting that:
$$ \frac{x^2}{a^2} = \cos^2 u, \; \; \; \frac{y^2}{c^2} = \sin^2 u\,\cos^2 v, \; \; \; \frac{z^2}{c^2} = \sin^2 u\,\sin^2 v $$
it follows that:
$$ \frac{x^2}{a^2} + \frac{y^2}{c^2} + \frac{z^2}{c^2} = 1 $$
which is the cartesian equation of a rotation ellipsoid, answer to the question.