I've been working through Silverman and Tate's book Rational Points on Elliptic Curves. They use conic equations as an introduction to singular/nonsingular curves. I've reproduced the problem with my comments below, then my questions follow.
Let $C$ be the conic given by the equation
$$F(x,y) = ax^2 + bxy + cy^2 + dx + ey + f = 0$$
and let $\delta$ be the determinant
$$\det \begin{bmatrix}
2a & b & d \\
b & 2c & e \\
d & e & 2f \\
\end{bmatrix}$$
(a) Show that if $\delta \neq 0$ then $C$ has no singular points.
Answered here.
(b) Conversely, show that if $\delta = 0$ and $b^2 – 4ac \neq 0$ then there is a unique singular point on $C$.
Given these two conditions we can classify the curve as two intersecting lines or a single point. So there is one singular point on the curve. This is from wikipedia.
(c) Let $L$ be the line $y = \alpha x + \beta$ with $\alpha \neq 0$. Show that the intersection of $L$ and $C$ consists of either zero, one, or two points.
Substitute the equation of the line into the equation of the conic. Then we get a quadratic equation $(a + \alpha b + \alpha^2 c)x^2 + (\beta b + 2\alpha\beta c + d + \alpha e)x + (c\beta^2 + \beta e + f) = 0$. This equation has zero, one, or two solutions.
(d) Determine the conditions on the coefficients which ensure that the intersection $L \cap C$ consists of exactly one point. What is the geometric significance of these conditions?
Let $(a + \alpha b + \alpha^2 c)x^2 + (\beta b + 2\alpha\beta c + d + \alpha e)x + (c\beta^2 + \beta e + f) = a'x^2 + b'x + c'$
There is exactly one solution if the determinant $b'^2 – 4a'c'$ of the quadratic equation is 0. Geometrically, the intersections collide: the line must be tangent.
My questions:
(b) Can anyone provide a reference other than Wikipedia for this? I've searched a lot but can't find anything that actually gives a derivation.
(c,d) What is the condition for zero, one, two roots? The field is not specified so I assume these roots could be complex. Then
- we have zero roots if $a' = b' = 0$
- we have one root if $a' = 0, b' \neq 0$
- we have two roots if $a' \neq, b' \neq 0$
But the two roots could be one root of multiplicity two. Is that correct? Or should I only be thinking about the third case (over real numbers), then considering the possible roots, with complex roots left out.
(d) How do I flesh out a proof of this fact? (i.e. How would I prove that $b'^2 – 4a'c'$ implies the line is tangent?)
Best Answer
b):
Let $Q(x,y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c$
$Q_x(x,y) = 2ax + 2hy + 2g$ and $Q_y(x,y) = 2hx + 2by+2f$
Solving $Q_x = 0$ and $Q_y = 0$, $$\begin{bmatrix}a &h\\ h & b \end{bmatrix}\begin{bmatrix}x\\ y \end{bmatrix} = \begin{bmatrix}-g\\ -f \end{bmatrix}$$
Given $ab - h^2 \ne 0$, there is a unique solution for $Q_x = 0, Q_y = 0$ and $Q$ has a unique centre. Let that centre be $(u ,v)$. Shifting the centre of $Q$ to origin with $x = X + u$ and $y = Y + v$ gives $$R(X, Y) = aX^2 + 2hXY + bY^2 + \dfrac{\Delta}{\gamma}$$
where $\Delta$ is determinant of matrix of $Q$ and $\gamma = ab - h^2$. Given $\Delta = 0$, origin lies on $R(X,Y)$. $$0 = R(0,0) = Q(u, v)$$
Therefore, there exist only one point $(u, v)$ for $Q_x = 0, Q_y = 0$ and $Q = 0$. Hence $(u, v)$ is unique singular point of $Q$.
c):
Let $L(t) = (u + tX, b + tY)$, putting this in $Q = 0$ give $pt^2 + qt + r = 0$ where
$$\begin{cases}p = aX^2 + 2hXY + bY^2 \\ q = X Q_x(u, v) + Y Q_y (u, v) \\ r = Q(u,v) \end{cases}$$
Hence $L$ intersects $Q$ at one point, two points or never provided $p, q, r$ all are not zero.
d): I don't understand this part. $L$ is tangent to $Q$ when $q^2 - 4pr = 0$ but finding the condition on coefficients of $Q$ and $L$ is a lot of dull calculations (at least with my method).