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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.