Confusion with the various forms of the equation of second degree I am confused with the second degree equation,an equation of second degree $ax^2+by^2+2hxy+2gx+2fy+c=0$ represents a conic,and nature of the conic depends on the various other conditions,like if $\Delta = abc + 2fgh - bg^2 - ch^2 - af^2 = 0$ and $h^2 \ge ab$ then this represent a pair of straight lines,$h^2 \gt ab$ then they are intersecting  straight lines, $h^2 = ab$ then parallel straight line .. like this other conditions for circle and ellipses are also given.
But I couldn't not understand how this is happening and no proof is given in my module and hence am not getting the feel of it,for circle and ellipse I could somehow see that if the substitute the equations appropriately I get the required equations of circle and ellipse or hyperbola but things are pretty much confusing for in case of pair of straight lines,my module stretches out the discussion on this by giving some formulas for the angle between the pair of straight lines,equation of the angle bisector then homogeneous form ($ax^2+2hxy+by^2$) area of the triangle formed by $ax^2+2hxy+by^2$ and $lx+my+n=0$ but all this things seems just like chug and plug formulas for me .. I don't want to memorize them just like that as I would forget them easily.
So could anybody suggest a proper reference in this regard?
 A: 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.
Analysis

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:
$$
\begin{cases}
  A &=& \cos^2(\phi) - \cos^2(\alpha)\cos^2(\gamma) \\ 
  B &=& - 2 \cos^2(\alpha)\cos(\gamma)\sin(\gamma) \\ 
  C &=& \cos^2(\phi) - \cos^2(\alpha)\sin^2(\gamma) \\ 
  D &=& 2 \left\{ \cos(\alpha)\cos(\gamma) (\vec{a}\cdot\vec{p}) 
        - \cos^2(\phi)\,p \right\} \\
  E &=& 2 \left\{ \cos(\alpha)\sin(\gamma) (\vec{a}\cdot\vec{p}) 
        - \cos^2(\phi)\,q \right\} \\
  F &=& (\vec{p}\cdot\vec{p}) \cos^2(\phi) - (\vec{a}\cdot\vec{p})^2
\end{cases}
$$
Meaning
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 $\overline{OA}=1$ .
A: Perhaps you'll find helpful this discussion from CRC Standard Mathematical Tables and Formulas. See also the Wikipedia page on degenerate conics.  


