I just started reading through Silverman and Tate's Rational Points on Elliptic Curves and I came across this passage.
Given a rational conic $\ldots$ suppose for now that we know of one rational point $O$ on our rational conic. Then we can get all of the rational points very simply. We just draw some rational line and project the conic onto the line from the point $O$ (to project $O$ itself onto the line, we use the tangent line to the conic at $O$).
A line meets a conic at two points, so for every point $P$ on the conic, we get a point $Q$ on the line. Conversely, for every point $Q$ on the line, by joining $Q$ to the point $O$, we get a point $P$ on the conic. In this way we get a one-to-one correspondence between points on the conic and points on the line.
Here, a rational conic is the set of roots $(x,y) \in \mathbb{R}^2$ of $ax^2 + by^2 + cxy + dx + ey + f$ where $a,b,c,d,e,f$ are all rationals. A rational point is a point with both coordinates rational. A rational line is a line that is the set of solutions to an equation of the form $ax + by + c = 0$ where $a,b,c$ are rational.
What I don't understand here is how to show that for every points $Q$ on the line, by joining $Q$ to point $O$, we get a point $P$ on the conic. This basically boils down to showing that a line that intersects a conic at exactly one point must be tangent to a conic. I tried writing down the equation for a line $px + qy +r = 0$ and the equation for a general conic (the equation above) and tried solving them. I got a quadratic, and the condition of intersection in one point is basically the condition that the discriminant of the quadratic is zero. But I don't see any way of bringing the tangency into this.
Is there a better way to do this? Or am I missing something?
As you say the common points of the line and the conic are the solutions of the system: $$ \begin{cases} ax^2 + by^2 + cxy + dx + ey + f=0\\ px+qy+r=0 \end{cases} $$ and, by substitution, this system becomes a second degree equation that has:
Two distinguished real solutions if the discriminant is $\Delta>0$, and this means that we have two distinct common points
No real solutions if the discriminat is $\Delta<0$ and this means that the stright line has no commn points with the conic
Only one real solution, that is a double root,if the discriminat is $\Delta=0$, and this means that the straight line has only one common point with the conic. This is exactly the caracterisation of the lines tangent to the conic, by definition.
If you use a definition of the tangent by means of the derivative of the function, note that the fact that the root is a double root, implies that the slope of the line is the value of the derivative at the tangence point.