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

Solution 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.

Solution 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 $(1,1)$.