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?