What is condition for second degree equation to represent a pair of straight lines?

Question

According to my text the necessary and sufficient condition for a general equation of second degree i.e. $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent a pair of straight lines is that 1) the discriminant $abc + 2fgh - af^2 - bg^2 - cf^2 = 0$ and 2) $h^2 \ge ab, g^2 \ge ca$ and $f^2 \ge bc$. I was able to prove part 1) but I am not able to work out why part 2) should always be satisfied. Can someone help prove part 2) and what happens if part 1) is true but part 2) is false?

I think $h^2 \ge ab$ must always be true because then angle between two lines represented by the equation of second degree will be not defined as $tan \theta = {2 \sqrt{h^2-ab} \over a+b} $ where $ \theta$ is angle between two lines.

Answer 1

If the conic represents a pair of lines, then it can be written as $(px+qy+r)(p'x+q'y+r')$. Therefore, we get $ab = pp'qq'$ and $h = \frac{pq'+p'q}{2}$.

1) If both $pq'$ and $p'q$ are greater than equal to zero then apply AM-GM to $pq'$ and $p'q$ to get

$$\frac{pq'+ p'q}{2} \geq \sqrt{pq' \cdot p'q}$$

Squaring both sides, we get

$$h^2 = {\bigg(\frac{pq'+ p'q}{2}\bigg)}^2 \geq pp'qq' = ab$$

2) If exactly one of $pq'$ and $p'q$ is less than zero then $ab \leq 0$ and therefore is always less than equal to $h^2 \geq 0$.

3) If both $pq'$ and $p'q$ are less than zero then apply AM-GM to $-pq'$ and $-p'q$ to get

$$\frac{-pq'+ (-p'q)}{2} \geq \sqrt{-pq' \cdot -p'q} = \sqrt{pp'qq'}$$

Squaring both sides, we get

$$h^2 = {\bigg(\frac{-pq'+ (-p'q)}{2}\bigg)}^2 \geq pp'qq' = ab$$

The other two conditions come out in a similar manner.

Answer 2

Hint:

The classification of conics is done through the quadratic form in $\mathbf R^3 $ associated to the matrix $$\begin{pmatrix} a&h&g\\h&b&f\\g&f&c \end{pmatrix}.$$ The conic splits into two lines if and only if the quadratic form has signature $(1,1)$.