Problem in Silverman/Tate Rational Points on Elliptic Curves I'm trying to figure out how to solve the following problem the "right" way. This is problem 1.2 on page 32:

Let $C$ be the conic given by the equation
$$F(x,y)=ax^2+bxy+cy^2+dx+ey+f = 0$$
Let $L$ be the line $y=\alpha x+\beta$ with $\alpha\neq 0$. Show that
  the intersection $L\cap C$ has zero, one or two points. Determine the
  conditions on the coefficients which ensure that the intersection
  $L\cap C$ consists of exactly one points. What is the geometric significance of these conditions?

The part about the number of points is trivial, since we just substitute the expression for $y$ and we are left with a quadratic in $x$.
After substituting the equation is
$$ax^2+b\alpha x^2+b\beta x + c\alpha^2x^2 + 2c\alpha\beta x + c\beta^2+e\alpha x+dx+e\beta+f = 0.$$
The coefficient of $x^2$ is simply $c\alpha^2+b\alpha+a$ which has discriminant $b^2-4ac$, so if $$\alpha = \frac{-b\pm \sqrt{b^2-4ac}}{2c},$$ then the equation becomes a linear one, so there's obviously only one solution. The problem is when $\alpha$ is not of this form. In that case it seems like one would have to compute the discriminant for the quadratic. The second condition would then be that the discriminant is zero.
I'm sort of curious what the problem is going after, since the discriminant being zero is clearly a condition on the coefficients. However, it's a pretty terrible expression, so I don't know how one could find any "geometric significance" in that expression?
Any ideas?

You want one point of intersection but with multiplicity 2. That is the the line and conic are tangent at that point. The way this works out in the algebra is that the quadratic equation you wrote should have both roots equal. 
The general line will meet the conic in two points. Some lines will not meet the conic (over the ground field) so the quadratic will not have  roots in that field but in an extension, for example if the field is the real numbers.
