Tangential space to the rational normal curve Exercise 15.5 (Harris, Algebraic Geometry: A First Course): Describe the tangential surface to the twisted cubic curve $C \subset \mathbb P^3$. In particular, show that it is a quartic surface. What is its singular locus?
Here the answer is given: $3x^2y^2-4x^3z-4y^3w+6xyzw-z^2w^2$ for $(w,x,y,z)=(1,t,t^2,t^3)$. Clearly, it is a quartic curve, and one can calculate its singular locus directly – it coincides with the twisted cubic curve itself.
But it does not explain anything. How to prove it without calculations? And what happens with the tangential space to the rational normal curve in $\mathbb P^n$ if $n>3$?
 A: It's not hard to determine parametric equations for the surface.  We have
$$x(t) = t, \qquad y(t) = t^2, \qquad z(t) = t^3,$$
so
$$x'(t) = 1, \qquad y'(t) = 2t, \qquad z'(t) = 3t^2;$$
therefore, the tangent line to a point $(t, t^2, t^3)$ is
$$(t, t^2, t^3) + s(1, 2t, 3t^2),$$
and consequently the surface $S$ consisting of all such lines is parametrized by
$$x(t,s) = t+ s, \qquad y(t,s) = t^2 + 2 s t, \qquad z(t,s) = t^3 + 3 s t^2.$$
(If you look at the answers to the question you linked, you can see that these equations were given as input to the computer.)
Implicitizing a set of parametric equations by hand is not straightforward.  However, the problem doesn't ask for an implicit equation; it just asks to indicate the surface (which we've done) and compute the singular locus.
But a having a parametrization instead of implicit equations actually makes it easier to determine where a surface is singular.  In particular,
$$\frac{\partial x}{\partial t} = 1, \qquad \frac{\partial x}{\partial s} = 1,$$
$$\frac{\partial y}{\partial t} = 2t + 2s, \qquad \frac{\partial y}{\partial s} = 2t,$$
$$\frac{\partial z}{\partial t} = 3t^2+6st, \qquad \frac{\partial z}{\partial s} = 3t^2,$$
so the tangent space to the surface $S$ at the point $\big(x(t,s), y(t,s), z(t,s)\big)$ is spanned by the vectors
$$\begin{bmatrix}
1 \\
2t + 2s \\
3t^2 + 6st \end{bmatrix} \qquad \text{ and } \qquad \begin{bmatrix}
1 \\
2t \\
3t^2\end{bmatrix}$$
By definition, $S$ is singular at $\big(x(s,t), y(s,t), z(s,t)\big)$ if the tangent space to $S$ at that point is less-than-two dimensional.  Clearly neither vector can be zero, so the only way this can happen is if one is a scalar multiple of the other.  Since both have the same first coordinate, this can only happen if they're equal.  And then it's easy to see that this happens only when $s = 0$, i.e. the singular locus of $S$ is the twisted cubic itself.
UPDATE: I see that the problem also asks to show that $S$ is a quartic surface.  If you have an appropriately general version of Bezout's theorem, you can do this by finding a line that meets $S$ transversally at exactly four points, though I don't immediately see a nice choice of line you could use.
A: Beltrametti, Carletti, Gallarati, Bragadin Lectures on Curves, Surfaces and Projective Varieties: A Classical View of Algebraic Geometry p 227-8:

Let $C$ be an irreducible non-planar cubic in ${\Bbb P}^3$ [..]for each point of ${\Bbb P}^3$ there passes one and only one chord (or a tangent) of $C$ , and write the equation for the ruled surface of the tangents to $C$ [..] :
For a point not belonging to a cubic $C$ there can not pass more than one chord of the curve, because otherwise $C$ would lie in the plane of two concurrent chords (since that plane has at least four intersections with $C$ ). The chords of $C$ (which are in number $\infty^2$) can not fill merely a surface, because in that case through each point of that surface there would have to pass infinitely many (i.e., the $\infty^1$ determined by passage through the point). On the other hand, a surface that contains all the chords of a curve would also have to contain all the cones that project the curve from one of its points, and the curve would be planar.
  This fact can be derived analytically in the following fashion. Let $C$ be the locus of the point $P(t) =[t^3,t^2,t,1]$ and $O = [a_0,a_1,a_2,a_3]$ a point not belonging to $C$. If $P(\lambda)$ and $P(\mu)$ are two points of the curve $C$ which are collinear with $O$, the matrix
\begin{pmatrix}a_0&a_1&a_2&a_3\\\lambda^3&\lambda^2&\lambda&1\\\mu^3&\mu^2&\mu&1\end{pmatrix}
has rank two. Hence
$$\begin{vmatrix}a_0&a_1&a_2\\\lambda^3&\lambda^2&\lambda\\\mu^3&\mu^2&\mu\end{vmatrix}=\begin{vmatrix}a_1&a_2&a_3\\\lambda^2&\lambda&1\\\mu^2&\mu&1\end{vmatrix}=0$$
That is,
  $$ a_0-a_1(\lambda+\mu)+a_2\lambda\mu=0 $$
  $$ a_1-a_2(\lambda+\mu)+a_3\lambda\mu=0$$
  This implies that  and  are the roots of the quadratic equation
$$\begin{vmatrix}x^2&x&1\\a_0&a_1&a_2\\a_1&a_2&a_3\end{vmatrix}=0 (*)$$
Thus for each point $O$ not belonging to $C$ there passes one and only one chord of $C$ . The trace $P'$ of this chord upon a plane $\pi$ not passing through $O$ is the double point of the cubic $\Gamma$, the projection of $C$ from $O$ onto $\pi$.
  If the roots of (*) coincide, that is, if O is a point of the quartic surface with equation
  $$(x_0x_3-x_1x_2)^2-4(x_0x_2-x_1^2)(x_1x_3-x_2^2)= 0 (**)$$ 
  one has a tangent rather than a chord of $C$ (at the point $P(\lambda)=P(\mu)$) and $P'$ is a cusp of $\Gamma$.
Equation (**) is the equation of the ruled surface F of the tangents of $C$ . From it one sees that a space cubic has rank 4, which means that four is the number of tangents of the cubic supported by a generic line (in the four points in which the line intersects the quartic $F$).

To what happens when $n > 3$: Eisenbud finds the quadrics in the ideal of the tangent developable surface to the rational normal curve (which generate the ideal from $n\geq 6$). 
He writes 

The following construction was noted by Buchweitz and Schreyer some time around 1983: One way to get a canonical curve of genus $g$ is to take a rational curve with $g$ cusps in ${\Bbb P}^{g-1}$. Such curves turn out to be the hyperplane sections of an arithmetically Cohen-Macaulay surface (actually a degenerate K3 surface, if there are any geometers listening) obtained as the tangent developable surface ($\equiv$ the union of the tangent lines) to the rational normal curve in ${\Bbb P}^g$

