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

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?

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 $$(x-p_1, y-p_2, z-p_3)$$)