Consider the projective $n$-space $\mathbb{P}^n$ over an algebraically closed field $k$ of characteristic $0$ (e.g. $k=\mathbb{C}$), and let $X$ be an integral closed subscheme of $\mathbb{P}^n$ of dimension $1$. For a nonsingular closed point $p\in X$, there is a unique line $L\subset\mathbb{P}^n$ such that $T_p(L)=T_p(X)$ as subspaces of $T_p(\mathbb{P}^n)$, and we say $L$ is the tangent line of $X$ at $p$. We say $X$ is a strange curve in $\mathbb{P^n}$, if there is a point $q\in\mathbb{P}^n$ such that for every nonsingular closed point $p$ of $X$, the tangent line of $X$ at $p$ passes through $q$. I want to show that, if $X$ is a strange curve in $\mathbb{P}^n$, then $X$ must be a line in $\mathbb{P}^n$, i.e., a complete intersection of $n-1$ hyperplanes.
This is the exercise IV.3.8(b) in R. Hartshorne's book Algebraic Geometry, and Hartshorne gave a proof for nonsingular case in theorem 3.9 of chapter IV. I first wanted to imitate the proof in theorem 3.9, but it seems that $X$ cannot always be embedded in $\mathbb{P}^3$, since there may exist bad singularity $p$ with (Zariski) tangent space of dimension $>3$.
Now I try to reduce this problem to the nonsingular case. My first attempt is to resolute the singularities of $X$ by normalization, but it does not preserve the "strange" property of curve (since it depends on the specified embedding $X\subset\mathbb{P}^n$). And my second attempt is to show that a strange curve has already been nonsingular (by e.g. showing it is of arithmetic genus $0$: using Hurwitz's formula for singular curves, I can show that the normalization of $X$ is of genus $0$ when considering the case $n=3$), but I have no idea to proceed and does not know how to use the assumption $\mathrm{char}\,k=0$.
The key here is using the fact that projection from a point which is on $T_pX$ for nonsingular $p$ makes $p$ a ramification point, i.e. $\Omega_{X/\pi(X)}$ has nonzero stalk at $p$. If you project from the point where all the tangent lines to the nonsingular points meet and the image $\pi(X)$ is a curve, you have a map of one-dimensional integral schemes over $k$ which is ramified at every nonsingular point. But this is impossible for a dominant map of one-dimensional integral schemes over $k$ in characteristic zero: the sheaf of relative differentials $\Omega_{X/\pi(X)}$ is coherent and zero at the generic point (because the stalk there is the module of differentials for the finite separable extension of function fields, hence zero) and therefore the support of $\Omega_{X/\pi(X)}$ must be a proper closed subset, i.e. finitely many points. So the projection of the curve must be a point, implying that the curve is a line.