Salmon's proof that tangents to a cubic from a point on it have the same cross-ratio In Higher plane curves, nr 167, Salmon proves that the cross-ration of the four tangents to a non-singular plane cubic, drawn from a point on the curve, is independent of the point. 
A proof can be found in Van der Waerden's Einführung in die algebraische Geometrie.
But can anyone explain in modern terms Salmon's proof? It runs as follows:
If from two consecutive points $O$, $O'$ of the curve we
draw the two sets of tangents $OA$, $OB$, $OC$, $OD$; $O'A$, $O'B$,
$O'C$, $O'D$, any tangent $OA$ intersects the consecutive tangent
$O'A$ in its point of contact. Now the four points of contact
$A$, $B$, $C$, $D$ lie on the polar conic of $0$, which also touches the
cubic at the point (Art. 64) ; hence the six points $OO'ABCD$
lie on the same conic, and therefore the anharmonic ratio of
the pencil $\{O.ABCD\}$ is the same as that of the pencil
$\{O'.ABCD\}$. Since then this ratio remains the same when we
pass from one point of the curve to the consecutive one, we learn
that the anharmonic ratio is constant of the pencil formed by the
four tangents which can be drawn from any point of the curve.
 A: The way I see this result is the following.
Fix a line $L \cong \mathbb{P}^1$ and a point $x \in X$, where $X \subset \mathbb{P}^2$ is the smooth plane cubic.
The projection $\pi_x \colon X \longrightarrow L$ is a double cover, branched in four points corresponding to the four tangent lines drawn from $x$ to $X$. The cross-ratio of these four points is equal to the cross ratio $\lambda(x)$ of the four tangents. On the other hand, the isomorphism class of the double cover is completely determined by the $j$-invariant $$j:= 2^8\frac{(\lambda^2- \lambda + 1)^3}{\lambda^2(\lambda-1)^2}.$$
Since $X$ is fixed, this means that the image of the holomorphic map  $$\lambda \colon X \longrightarrow \mathbb{C}, \quad x \mapsto \lambda(x)$$ 
is a finite set of values. But then $\lambda$ is constant.
A: Consider a local parametrization $O(t), A(t), ...$, with $O = O(0), A = A(0), ...$. Salmon is essentially checking that $\frac{d}{dt}(O(t)A(t),O(t)B(t);O(t)C(t),O(t)D(t)) = 0$ at $t = 0$, where $(OA,OB;OC,OD)$ denotes the anharmonic ratio of the pencil $\{O.ABCD\}$.
Since $A(t)$ stays on the line $OA$ to first order, and similarly for $B(t), C(t), D(t)$, we have
$\frac{d}{dt}(O(t)A(t),O(t)B(t);O(t)C(t),O(t)D(t))|_{t=0} = \frac{d}{dt}(O(t)A,O(t)B;O(t)C,O(t)D)|_{t=0}$.
This will be zero if and only if $O(t)$ stays on the conic $OABCD$ to first order, which Salmon checks using the fact that for $O$ on the cubic curve the polar conic of $O$ passes through $O, A, B, C, D$ and is tangent to the cubic curve at $O$.
