Is this textbook formula for line-curve intersections missing a "general position" assumption? (Dragović & Radnović)

I am working through Poncelet Porisms and Beyond by V. Dragović and M. Radnović. In Chapter 3 (Section 3.2, "Algebraic curves in the complex projective plane"), I am stuck on an exercise that makes two claims about the intersection of lines with an algebraic curve.

Here is the exact quote from the book:

Exercise 3.32. Let $p(x, y) = 0$ be an algebraic curve of degree $d$ in $\mathbb{C}^2$. Prove that, for each point $(x_0, y_0)$, the following equality holds: $$\sum_{(\alpha : \beta) \in \mathbb{C}P^1} \left(d - \deg p(x_0 + \alpha t, y_0 + \beta t)\right) = d,$$ where the sum is taken over the set of all lines containing $(x_0, y_0)$.

Moreover, each member of the sum, i.e., the degree of the polynomial $p(x_0 + \alpha t, y_0 + \beta t)$, does not depend on point $(x_0, y_0)$, but only on pair $(\alpha, \beta)$.

When trying to prove this, I tested the statements on simple conics and found that both claims seem to fail. Here is my work:

1. Testing the sum formula (Counterexample to the first claim) Consider a parabola $p(x, y) = y^2 - x = 0$ (so $d = 2$). Let's choose the origin $(x_0, y_0) = (0, 0)$. The line parametrization is $x = \alpha t, y = \beta t$. The restricted polynomial is: $$Q_{(\alpha, \beta)}(t) = p(\alpha t, \beta t) = \beta^2 t^2 - \alpha t$$

• If $\beta \neq 0$ (all non-horizontal lines), the coefficient of $t^2$ is non-zero. Thus $\deg Q = 2$, and the term in the sum is $d - \deg Q = 2 - 2 = 0$.
• If $\beta = 0$ (the horizontal line $y = 0$, so $\alpha \neq 0$), the equation is $Q(t) = -\alpha t$. Here $\deg Q = 1$, and the term in the sum is $2 - 1 = 1$.

Summing over all directions $(\alpha : \beta) \in \mathbb{C}P^1$, I get: $$\sum (d - \deg Q) = 1 + 0 = 1.$$ But the formula claims the sum should be exactly $d = 2$. (I suspect this happens because the parabola is tangent to the line at infinity, meaning intersection multiplicity is lost, but the text doesn't mention this).

2. Testing the independence of the base point (Counterexample to the second claim) Consider the hyperbola $p(x, y) = x^2 - y^2 - 1 = 0$ ($d = 2$), and fix the asymptotic direction $(\alpha : \beta) = (1 : 1)$. Let's calculate the degree of the polynomial from two different base points:

• From $(x_0, y_0) = (0, 0)$: $$p(t, t) = t^2 - t^2 - 1 = -1 \implies \deg Q = 0.$$
• From $(x_0, y_0) = (1, 0)$: $$p(1 + t, t) = (1 + t)^2 - t^2 - 1 = 2t \implies \deg Q = 1.$$

The degree explicitly changes depending on the choice of $(x_0, y_0)$, which directly contradicts the second paragraph of the exercise.

My Questions:

1. Am I misunderstanding the standard definitions of polynomial degree or line parametrization in this context, or is the exercise mathematically flawed as written?
2. If it is flawed, what are the exact implicit assumptions (e.g., "the curve must be in general position with respect to the line at infinity without multiple roots") required to make this exercise correct?

Any clarification on how to properly rigorously formulate this theorem would be greatly appreciated!