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.
Here's a graphical, hand-wavy, interpretation:
$\Phi(x,y)=0$ is the implicit function of a certain graph in the $(x,y)$ plane. We know that: $$\frac{dy}{dx}=-\frac{\dfrac{\partial \Phi}{\partial x}}{\dfrac{\partial \Phi}{\partial y}}$$
If $\dfrac{\partial \Phi}{\partial x}=0$ and $\dfrac{\partial \Phi}{\partial y}=0$ then $dy/dx$ is indeterminate, and the only place that's true for two lines is their intersection.
Best Answer
Here are two ways to answer the question.
It is clear that the following condition must hold:
$$\tag{0}m:=a b - h^2 \neq 0$$
(otherwise, the LHS of the given equation would become a constant).
The equation of the conical section can thus be divided by $m$, and therefore be written
$$\tag{1}U:=ax^2+2hxy+by^2+2gx+2fy+c=0 \ \ \text{where} \ \ c:=\frac{a f^2 + b g^2 - 2f g h}{a b - h^2}$$
or, in a matrix form:
$U=X^TMX=0$ with $X^T=(x \ y \ 1)$ and:
$$\tag{2} M=\pmatrix{a& h& g\\h& b& f\\g& f& c}.$$
Expanding the determinant of $M$ along its third column gives $det(M)=0$.
Therefore $rank(M)\leq 2$.
But condition (0) expresses the fact that the upper left minor of $M$ is nonzero. Thus $rank(M)\geq 2$. Finally:
$$\tag{3} rank(M)=2.$$
Thus the conical section is decomposable into 2 straight lines (see for example (Decomposition of a degenerate conic)), but these straight lines might have complex coefficients!
It appears that a condition of reality is
$$\tag{3}h^2 > ab$$
(were you aware of it?) as we are going to see it in the second way.
Under condition (3), (1) can be written
$$\tag{4}U=\left(s x + \frac{h y + g}{s}\right)^2-\left(t y + \frac{u}{t}\right)^2=0$$
by setting $s=\sqrt{a}, t=\dfrac{\sqrt{h^2 - a b}}{s}, u=\dfrac{g h - a f}{a}.$
Being a difference of two squares, (4) can be decomposed into two pairs of first degree equations which are the equations of the straight lines.