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Find the tangent space of $$M = \{(x,y,z)|\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\}$$

So I know the formula of tangent space for a manifold represnted by $F$ such that $F=0$: it is $ker (DF)$.

So I'll define - $F = \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} - 1$ and of course $F=0$.

By definition, $DF = (\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2})$ and we just need to find $ker (DF)$.

Besides $x=y=z=0$, the solutions are $(x,y,(-\frac{x^2}{a^2} -\frac{y^2}{b^2} )c^2)$, $(x,(-\frac{x^2}{a^2} -\frac{z^2}{c^2})b^2,z)$ and $(-\frac{y^2}{b^2} -\frac{z^2}{c^2} )a^2,y,z)$.

But what is the final tangent space that is spanned by these solutions?

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At a point $p \in M$, the tangent space to $M$ is given by $ker(DF(p))$, as you said. So, in your case, it is the set of points $(x,y,z) \in \mathbb{R^3}$ such that (I divided the $2$ which comes from differentiating)

\begin{align*} \begin{pmatrix} \dfrac{p_1}{a^2} & \dfrac{p_2}{b^2} & \dfrac{p_3}{c^2} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= 0 \\ \dfrac{p_1}{a^2}x + \dfrac{p_2}{b^2}y + \dfrac{p_3}{c^2}z &= 0 \end{align*}

In other words the tangent space of $M$ at $p$ is the plane through the origin given by the equation above.

(If you want the actual tangent plane to $M$ at the point $p$, you simply have to translate the plane to pass through $p$, by replacing $(x,y,z)$ with $(x-p_1, y-p_2, z-p_3)$)

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