Tangent space of a line

Question

The set of all lines passing through the origin:$$L=\{(x,y)\mid y=ax, a \in \mathbb{R}\}$$ I want to compute $T_{p}L$, I also know that $T_{p}\mathbb{R^{2}}=Sp\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\}$. It is clear that $L \subset \mathbb{R^{2}}$, so every $v \in T_{p}L$ can be written as a linear combination of $\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\}$, $$v=\alpha\frac{\partial}{\partial x} + \beta\frac{\partial}{\partial y}$$ From this step I don't know how to continue.

Edit

$L$ is considered as a submanifold of $\mathbb{R^{2}}$

Answer 1

Hint: Take the linear function $\ell(x,y)=y-ax$ and compute the partial derivatives $\frac{\partial \ell}{\partial x}$ and $\frac{\partial \ell}{\partial y}$.

Answer 2

Strictly speaking, $L$ doesn't consist of all lines in the plane through the origin. It's missing part of the vertical "$x=0$" line. But I think it's easy enough to see that $L$ is an open subset of $\Bbb{R}^2$, hence its tangent space can be identified with that of the whole plane: \begin{align} T_p(L) = T_p(\Bbb{R}^2). \end{align}

This is a special case of the general fact that if $U$ is an open subset of a smooth manifold $M$, then $T_pU = T_pM$; or atleast $T_pU$ is canonically isomorphic to $T_pM$ (depending on your definition of tangent space).

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Warning:

This answer is only if you really meant the set of "all" lines through the origin, as opposed to having a fixed value $a \in \Bbb{R}$, and considering the single line $\{(x,y) \in \Bbb{R}^2| \, y = ax\}$. If you meant only a single line, then you should refer to the hint given by Wuestenfux.