Show that it is a smooth surface

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Show that the ellipsoid $$\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1$$ where $p$, $q$ and $r$ are non-zero constants, is a smooth surface.

To do this do we have to take a parametrization of the ellipsoid?

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You don't have to simultaneously parameterize the entire ellipsoid. It works just as well to construct two or three or more parameterized charts on the ellipsoid, as long as the union of those charts equals the entire ellipsoid.

So, for example, by solving for $z$ you get two parameterized charts: $$z = \pm \sqrt{r^2 - \frac{r^2}{p^2} x^2 - \frac{r^2}{q^2} y^2} $$ where the $+$ sign is used for the chart where $z>0$ and the $-$ sign is used for the chart where $z<0$. The domain of the parameterization is the open subset $U$ of the $(x,y)$ plane where the stuff under the $\sqrt{}$ sign is positive (geometrically it is the interior of an ellipse). So, for example, the chart where $z>0$ is described in full detail by the formula $$(x,y) \mapsto \biggl(x,y,\sqrt{r^2 - \frac{r^2}{p^2} x^2 - \frac{r^2}{q^2} y^2}\biggr), \quad (x,y) \in U $$ As you can see, this chart is defined as a graph of a function, namely $z$ as a function of $(x,y)$.

By solving for $x$ you get two more charts, each expressing $x$ as a function of $(y,z)$, one for $x>0$ and the other for $x<0$.

And by solving for $y$ you get still another two charts, each expressing $y$ as a function of $(z,x)$, one for $y>0$ and the other for $y<0$.

Since each point on the ellipse has at least one nonzero coordinate, these six charts cover the whole ellipsoid. And since each chart is parameterized by a smooth function, you may conclude that the ellipsoid is a smooth surface.