Why does this formula from Silverman give the tangent lines of a cusp or node? On page 44 of Arithmetic of ECs, Silverman explains that given a weierstrass equation $f \in K[x,y]$ and singular point $P=(x_0, y_0)$ with $\partial f / \partial x(P) = \partial f / \partial y (P) = 0$ that one can rearrange the Taylor expansion of $f$ at P to get: $$f(x,y) - f(x_0, y_0) = ((y-y_0) - \alpha (x-x_0))((y-y_0) - \beta (x-x_0)) - (x- x_0)^3,$$ for some $\alpha, \beta \in \overline K$. He then goes on to say $(y-y_0) - \alpha (x-x_0)$ and $(y-y_0) - \beta (x-x_0)$ are the tangent lines of $f$ at $P$ (if $P$ is a cusp then they coincide). 
This might be a silly question, but why are these the tangent lines? I assume this follows from where they appear in the Taylor expansion but I can't figure out why.
 A: Let $C$ be an algebraic curve given by $f(x,y)=0$, where $f$ is a polynomial in two variables. If $C$ is non-singular at $P=(x_0,y_0)$, then we can "linearize" $C$ at $P$ using the tangent line $$L : (\partial f/\partial x)|_P \cdot (x-x_0) + (\partial f/\partial y)|_P \cdot (y-y_0) = 0,$$
which is the 1st degree Taylor polynomial in two variables (note that $P\in C$ so $f(x_0,y_0)=0$ here). Similarly, we can find the "best quadratic approximation" (or best conic) of $C$ at $P$ by using the 2nd degree Taylor approximation:
$$Q : (\partial f/\partial x)|_P \cdot (x-x_0) + (\partial f/\partial y)|_P \cdot (y-y_0)+ \frac{1}{2}(\partial^2 f/\partial x^2)|_P \cdot (x-x_0)^2+(\partial^2 f/\partial x \partial y)|_P \cdot (x-x_0)(y-y_0) + \frac{1}{2}(\partial^2 f/\partial y^2)|_P \cdot (y-y_0)^2 = 0.$$
If the curve $C$ is singular at $P$, then both first derivatives vanish and the linear approximation gives no information. However, the second degree approximation becomes:
$$Q' :  \frac{1}{2}(\partial^2 f/\partial x^2)|_P \cdot (x-x_0)^2+(\partial^2 f/\partial x \partial y)|_P \cdot (x-x_0)(y-y_0) + \frac{1}{2}(\partial^2 f/\partial y^2)|_P \cdot (y-y_0)^2 = 0,$$
and this must be a singular conic (i.e., the product of two lines). Hence, the equation defining $Q'$ must factor as $(ax+by+c)(dx+ey+f)=0$.
What Silverman is doing is writing the 3rd degree Taylor polynomial of $f(x,y)$. Since the equation is cubic, the 3rd degree Taylor polynomial is identical to $f(x,y)$. But once we have identified the Taylor expression, we can isolate the second degree polynomial also, and the quadratic terms must correspond to the expression I wrote for $Q'$ above, and since we have a singularity, it must factor as a product of two lines, which are the lines he mentions. 
