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?


2 Answers 2


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.
enter image description here
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: $$ \begin{cases} \mbox{Circle : }& \alpha = 90^o &\quad \Longleftrightarrow \quad \epsilon = 0 \\ \mbox{Ellipse : }& \alpha > \phi &\quad \Longleftrightarrow \quad \epsilon < 1 \\ \mbox{Parabola : }& \alpha = \phi &\quad \Longleftrightarrow \quad \epsilon = 1 \\ \mbox{Hyperbola : }& \alpha < \phi &\quad \Longleftrightarrow \quad \epsilon > 1 \end{cases} $$ enter image description here

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 $$ \vec{a} = \left[ \cos(\alpha)\cos(\gamma),\cos(\alpha)\sin(\gamma),\sin(\alpha)\right] $$ enter image description here
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 $\overline{OA}=1$ .

  • $\begingroup$ Your answer is excellent but I couldnot follow it until last.I was stumped at how you expressed $\vec{a}$ in terms of various angles in the first para under Analysis section.If its possible please give a bit more explanation otherwise I might have to wait till i become more mathematically mature.Thanks. $\endgroup$
    – Navin
    Commented Jan 27, 2017 at 11:18
  • $\begingroup$ @navinstudent: I've made an EDIT to the answer. Hope it helps. $\endgroup$ Commented Jan 27, 2017 at 13:36
  • $\begingroup$ @navinstudent This answer to a more recent question about quadratic equations of two variables may be relevant: math.stackexchange.com/questions/2096193/… -- but at the time I wrote that, I was not aware of this question, and classification of the shapes of the curves was a byproduct of the answer rather than a goal. $\endgroup$
    – David K
    Commented Jan 27, 2017 at 14:27
  • $\begingroup$ Your edit was helpful and i could follow it till last.Thanks a lot. $\endgroup$
    – Navin
    Commented Jan 27, 2017 at 17:30

Perhaps you'll find helpful this discussion from CRC Standard Mathematical Tables and Formulas. See also the Wikipedia page on degenerate conics.

enter image description here enter image description here

  • $\begingroup$ Thank you I also took the help of this page. $\endgroup$
    – Quixotic
    Commented Apr 21, 2011 at 19:40

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