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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.

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  • $\begingroup$ With $h^2<ab$, the trinomial formed by the quadratic terms is positive (or negative) definite, hence the conic is of the ellipse type. $\endgroup$ – Yves Daoust Apr 14 '16 at 20:38
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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.

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  • $\begingroup$ thanks that was pretty easy.! $\endgroup$ – Matt Apr 14 '16 at 21:04
  • $\begingroup$ I realized that we can't quite apply AM-GM since $a$ and $b$ need not be positive in general. But in that case, $ab \leq 0$ and is hence less than equal to the square $h^2$. $\endgroup$ – Seven Apr 14 '16 at 21:11
  • $\begingroup$ may you please show how can we prove any one of three by A.M G.M inequality? $\endgroup$ – Matt Apr 15 '16 at 17:11
  • $\begingroup$ @Raghav I have elaborated on the answer. Hope that helps. $\endgroup$ – Seven Apr 15 '16 at 20:36
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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 $(1,1)$.

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