conic sections
For the standard conic equation in 2-D plane i.e
$$ax^2 + 2hxy + by^2 +2gx +2fy +c =0$$
If $a=b=0$ and $h\neq 0$, then the equation reduces to
$$2hxy + 2gx + 2fy + c =0$$
Under what conditions will the above equation represent a pair of straight lines?
The theory of Conic Sections stems from ancient times. It is an example of pure mathematics, which has found applications only many centuries after it
has been developed, e.g. with the laws of planet motion as discovered
by Johannes Keppler. But, interesting as it is, we shall leave aside
history and immediately come to core - or rather
cone - business.
The problem is to intersect a
circular cone with a plane and determine the curves of intersection, like shown
in the above picture. This could be done in the way the old Greek mathematicians
did it. But we prefer to take a path that requires less ingenuity and we shall
employ the means of modern analytical geometry instead. With such an approach,
though, one should be prepared for tedious algebra when working out
the details.
A circular cone is characterized by the fact that the angle $\phi$ between the cone axis and its surface is a constant. Let the unit vector $\vec{a}$ be the direction of the cone axis and let $\vec{p}$ point to the top vertex of the cone. An arbitrary point at the surface of the cone is pinpointed by $\vec{r}$. Then the following is an equation of the cone surface: $$ (\vec{a}\cdot\vec{r}-\vec{p}) = |\vec{a}||\vec{r}-\vec{p}|\cos(\phi) $$ Square both sides: $$ (\vec{a}\cdot\vec{r}-\vec{p})^2 = (\vec{a}\cdot\vec{a})(\vec{r}-\vec{p}\cdot\vec{r}-\vec{p})\cos^2(\phi) $$ And work out: $$ (\vec{a}\cdot\vec{r})^2 - 2(\vec{a}\cdot\vec{p})(\vec{a}\cdot\vec{r}) + (\vec{a}\cdot\vec{p})^2 = \cos^2(\phi) \left\{ (\vec{r}\cdot\vec{r}) - 2(\vec{p}\cdot\vec{r}) + (\vec{p}\cdot\vec{p})\right\} $$ The unit vector $\vec{a}$ can be written as: $$ \vec{a} = \left[ \cos(\alpha)\cos(\gamma),\cos(\alpha)\sin(\gamma),\sin(\alpha)\right] $$ Where $\alpha$ is the angle between the cone axis and the XY-plane and $\gamma$ is an angle that indicates how the conic section is rotated in the plane. The vector of the top of the cone can be written in its coordinates as: $$ \vec{p} = (p,q,h) $$ Where $h$ is the height of the cone above the XY plane and $(p,q)$ indicates how the conic section is translated in the plane. Last but not least, the vector pointing to the cone surface is written as: $$ \vec{r} = (x,y,z) $$ Where the intersections with the XY plane are found for $z = 0$. Let's do just that and work out the above: $$ \begin{cases} (\vec{a}\cdot\vec{r}) &=& \cos(\alpha)\cos(\gamma)\,x + \cos(\alpha)\sin(\gamma)\,y \\ (\vec{a}\cdot\vec{p}) &=& \cos(\alpha)\cos(\gamma)\,p + \cos(\alpha)\sin(\gamma)\,q + \sin(\alpha)\,h \\ (\vec{r}\cdot\vec{r}) &=& x^2 + y^2 \\ (\vec{p}\cdot\vec{r}) &=& p\,x + q\,y \\ (\vec{p}\cdot\vec{p}) &=& p^2 + q^2 + h^2 \end{cases} $$ Collecting powers of $x$ and $y$ results in: $$ A\,x^2 + B\,xy + C\,y^2 + D\,x + E\,y + F = 0 $$ Where: $$ \begin{cases} A &=& \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \\ B &=& - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \\ C &=& \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma) \\ D &=& 2 \left\{ \cos(\alpha)\cos(\gamma) (\vec{a}\cdot\vec{p}) - \cos^2(\phi)\,p \right\} \\ E &=& 2 \left\{ \cos(\alpha)\sin(\gamma) (\vec{a}\cdot\vec{p}) - \cos^2(\phi)\,q \right\} \\ F &=& (\vec{p}\cdot\vec{p}) \cos^2(\phi) - (\vec{a}\cdot\vec{p})^2 \end{cases} $$
The first three coefficients of the conic section equation are:
$$
\begin{cases}
A &=& \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \\
B &=& - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \\
C &=& \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma)
\end{cases}
$$
All kind of conics can still be produced if the angles $\phi$ and $\alpha$ are
limited to sensible values:
$$
\begin{cases}
&&0 < \phi < 90^o \quad \Longrightarrow \quad 0 < \cos(\phi) < 1 \\
&&0 \le \alpha \le 90^o \quad \Longrightarrow \quad 0 \le \cos(\alpha) \le 1
\end{cases}
$$
Generality is not affected by these choices. Moreover it is seen from the picture
below that the form of the conic section is determined
by the angles $\phi$ and $\alpha$ and nothing else. Therefore the ratio of the
two angles will be defined here as the excentricity ($\epsilon$) of the
conic section:
$$
\epsilon = \frac{\cos(\alpha)}{\cos(\phi)}
$$
The following relationships exist between the excentricity and the form of a
conic section, as is clear from the picture:
$$
\begin{cases}
\mbox{Circle : }& \alpha = 90^o &\quad \Longleftrightarrow \quad \epsilon = 0 \\
\mbox{Ellipse : }& \alpha > \phi &\quad \Longleftrightarrow \quad \epsilon < 1 \\
\mbox{Parabola : }& \alpha = \phi &\quad \Longleftrightarrow \quad \epsilon = 1 \\
\mbox{Hyperbola : }& \alpha < \phi &\quad \Longleftrightarrow \quad \epsilon > 1
\end{cases}
$$
So far so good. The coefficients $(A,B,C)$ can be combined into
some interesting quantities which are only dependent upon form,
that is: the angles $\phi$ and $\alpha$.
It is remarked in the first place that $(A,B,C)$ are independent of the vector
$\vec{p} = (p,q,h)$ and thus independent of translation and scaling. If we seek
to eliminate any dependence upon the angle of rotation $\gamma$, then we find:
$$
A + C = 2 \cos^2(\phi) - \cos^2(\alpha) = \cos^2(\phi) (2 - \epsilon^2)
$$
This quantity is known (for some good reasons) as the trace of the conic
section. Instead of eliminating the angle of rotation, we could also try
to calculate it.
$$
A - C = - \cos^2(\alpha)\left[\cos^2(\gamma)-\sin^2(\gamma)\right]
= - \cos^2(\alpha)\cos(2\gamma)
$$
There is a striking resemblance with:
$$
B = - \cos^2(\alpha)\sin(2\gamma)
$$
We thus find:
$$
\frac{B}{A - C} = \frac{\sin(2\gamma)}{\cos(2\gamma)} \quad \Longrightarrow \quad
\tan{2\gamma} = \frac{B}{A - C}
$$
Herewith - in principle - the angle of rotation $\gamma$ can be reconstructed
from the conic section equation; provided that $A \neq C$.
Let's proceed with
another quantity that is independent of any rotation.
$$
B^2 - 4 A C = \left[ - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \right]^2
$$ $$
- 4 \left[ \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \right]
\left[ \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma) \right]
$$
This quantity is known (also for some good reasons) as the determinant or
discriminant of the conic section. Work out:
$$
= 4 \cos^4(\alpha)\cos^2(\gamma)\sin^2(\gamma) - 4 \cos^4(\phi)
- 4 \cos^4(\alpha)\cos^2(\gamma)\sin^2(\gamma) \\
+ 4 \cos^2(\phi)\cos^2(\alpha)\left[\cos^2(\gamma)+\sin^2(\gamma)\right] \\
= - 4\cos^4(\phi) + 4\cos^2(\phi)\cos^2(\alpha) \\
\quad \Longrightarrow \quad B^2 - 4 A C
= 4\cos^2(\phi)\left[\cos^2(\alpha) - \cos^2(\phi)\right]
= \left[2\cos^2(\phi)\right]^2(\epsilon^2 - 1)
$$
The following relationships exist between the discriminant and the form of a
conic section, as is clear from the above:
$$
\begin{cases}
&\mbox{Ellipse}& \quad \Longleftrightarrow \quad \epsilon < 1 \quad \Longleftrightarrow \quad (B^2 - 4 A C) < 0 \\
&\mbox{Parabola}& \quad \Longleftrightarrow \quad \epsilon = 1 \quad \Longleftrightarrow \quad (B^2 - 4 A C) = 0 \\
&\mbox{Hyperbola}& \quad \Longleftrightarrow \quad \epsilon > 1 \quad \Longleftrightarrow \quad (B^2 - 4 A C) > 0
\end{cases}
$$
BONUS on excentricity. $$ (A + C)^2 = 4\cos^2(\phi)\left[\cos^2(\phi) - \cos^2(\alpha)\right] + \cos^4(\alpha) $$ Upon addition this gives: $$ (B^2 - 4 A C) + (A + C)^2 = \cos^4(\alpha) \quad \Longrightarrow \\ \cos(\alpha) = \sqrt{\sqrt{B^2 + (A-C)^2}} $$ About the angle $\phi$ between the cone axis and its surface: $$ A + C = 2\cos^2(\phi) - \sqrt{B^2 + (A-C)^2} \quad \Longrightarrow \quad \cos(\phi) = \sqrt{\frac{(A+C) + \sqrt{B^2 + (A-C)^2}}{2}} $$ Herewith the excentricity $\epsilon$ can be expressed into the coefficients of the conic section equation $(A,B,C)$: $$ \epsilon = \sqrt{\frac{2\sqrt{B^2 + (A-C)^2}} {(A+C) + \sqrt{B^2 + (A-C)^2}}} $$ Many other expressions can be derived, especially for the coefficients $D,E,F$. But I think that deriving them here will disturb a good balance between interesting and cumbersome.
EDIT. Explanation of
$$
\vec{a} = \left[
\cos(\alpha)\cos(\gamma),\cos(\alpha)\sin(\gamma),\sin(\alpha)\right]
$$
See picture. Overline denotes " length of " in:
$$
\overline{OZ}=\overline{OA}\, \sin(\alpha) \quad ; \quad
\overline{OD}=\overline{OA}\, \cos(\alpha) \\
\overline{OX}=\overline{OD}\, \cos(\gamma) \quad ; \quad
\overline{OY}=\overline{OD}\, \sin(\gamma)
$$
With $\overline{OA}=1$ .
Dividing $f(x,y)$ through by the suggested $x+2y+3$ gives $$f(x,y) = (x+2y+3)(2x-3y-4)=0.$$ The product is zero when either $x+2y+3=0$ or $2x-3y-4=0$, both of which are equations for lines.
You're right that $f$ is has positive discriminant, but it happens to be a reducible degenerate conic. Maybe the simplest example is $y^2-x^2=0$, which is clearly a pair of lines. Generally speaking, a conic section $f(x,y)=0$ will be degenerate any time you can factor $f(x,y) = a(x,y)b(x,y)$.
Best Answer
The equation $2hxy + 2gx + 2fy + c =0$ is expected to be factorized as
$$ (px+q)(ty+s)=0$$
in order for it to present two lines. Match the coefficients to get,
$$pt = 2h, \>\>\> qt = 2g,\>\>\> sp=2f, \>\>\> qs = c$$
Then,
$$\frac pq =\frac hg = \frac {2f}c$$
Thus, the condition is $hc = 2fg$.