# How to determine standard equation of a conic from the general second degree equation?

From a given general equation of second degree i can determine the conic by following rules:

Given equation: $ax^2+by^2+2hxy+2gx+2fy+c=0$

then if,

$abc+2fgh-af^2-bg^2-ch^2$ is not equal to zero

the equation represents:

• Parabola if $h^2=ab$
• Ellipse if $h^2<ab$
• Hyperbola if $h^2>ab$

However i am unable to determine the conic in standard form. For the conics that have axes parallel to coordinate axes, i can use completing the square method. But it becomes very difficult when it comes to determining the equation of an inclined conic. Is there any other method to determine the standard equation?

The method to determine it for a parabola is here but how to the same for an ellipse and hyperbola?

• This book should be useful. I recommend it. – Giuseppe Negro May 3 '16 at 7:23
• Have you studied basic linear algebra: matrices, determinants and etc.? Because then it is pretty easy to classify two and three dimensional conics. – DonAntonio May 3 '16 at 7:31
• First get rid of the $xy$ term. Do a rotation, as in the reference you give in your question. Then, in the new reference frame, your conic is parallel to the axes – Andrei May 3 '16 at 7:34
• @Joanpemo I have studied basics of matrices, determinants etc. so go ahead with your answer, their use for determining equation seems interesting. – James Hunt May 3 '16 at 8:04
• @Andrei I know the procedure of rotation for equations of first degree curves(lines) only. How to apply the same for a second degree equation? – James Hunt May 3 '16 at 8:05

The "complete the square" method allows you to center the conic, i.e. make the linear terms vanish (by a translation, $ax^2+bx+c\to a'u^2+c'$).

To deal with "obliqueness", you need to let the cross term $xy$ vanish, by means of a rotation.

Let $x=cu-sv,y=su+cv$, which expresses a rotation around the origin ($c,s$ denote the cosine and sine of the angle), and let the centered equation be in the form

$$Ax^2+2Bxy+Cy^2=1.$$

Then substituting,

$$(Ac^2+2Bcs+Cs^2)u^2+(-2Acs+B(c^2-s^2)+2Ccs)uv+(As^2-2Bcs+Cc^2)v^2=1.$$

By a suitable choice of the angle, you can achieve

$$-2Acs+B(c^2-s^2)+2Ccs=0$$

and the equation reduces to

$$A'u^2+C'v^2=1.$$

Depending on the signs of $A',C'$, you get an ellipse or an hyperbola. By a further rescaling of the variables, you can obtain the "canonical" forms (circle and equilateral hyperbola)

$$p^2\pm q^2=1.$$

To find the suitable angle, use the "double angle" formulas and rewrite

$$(C-A)\sin(2\theta)+B\cos(2\theta)=0,$$ which is easy to solve.

• You are a wizard! where did you learn this wizardry from? – jonsno Apr 3 '17 at 11:51
• @samjoe: this is classical theory of conics and in close relation to linear algebra. The rotation trick is actually a matrix diagonalization. – Yves Daoust Apr 3 '17 at 11:53
• @YvesDaoust Who is samjoe here? – user91594 Dec 23 '20 at 6:13
• @user91594: probably someone who discarded his comment. Does this matter ? – Yves Daoust Dec 23 '20 at 14:37
• @YvesDaoust well i found out that jonsno(John Snow) is the samjoe(mystery solved). For a future reader – Anonymous Dec 25 '20 at 11:27

We have $\;ax^2+by^2+2hxy+2gx+2fy+c=0\;$ . Let us write down this quadratic's expanded matrix:

$$A:=\begin{pmatrix}c&g&f\\g&a&h\\f&h&b\end{pmatrix}$$

Observe that firs row is [free coefficient , half coefficient of x, half coefficient of y], the second row is [hlaf coef. of x, coeff. of x^2 , half coef of xy] and etc.

The corresponding reduced matrix is

$$A':=\begin{pmatrix}a&h\\h&b\end{pmatrix}$$

We have, of course, that both matrices are symmetric. Let us now define for a square matrix $\;B\;$ :

$$\color{red}{r(B)}:=\text{ the rank of}\;\;B\;,\;\;\color{red}{|B|}:=\det B\;,\;\;\color{red}{\sigma(B)}:=\text{ the signature of}\;\;B$$

where signature = the difference between the number of positive elements with the number of negative elements in the main diagonal of the matrix.

Then we have, for example (the whole table is way too long):

$$\begin{cases}r(A)=3\;,\;\;r(A')=2\;,\;\;|A|,A'|>0\;\color{red}\implies\;\text{Empty set}\;(\emptyset)\\{}\\r(A)=3\;,\;\;r(A')=2\;,\;\;|A|<0,\;|A'|>0\;\color{red}\implies\;\text{Ellipse}\\{}\\r(A)=3\;,\;\;r(A')=2\;,\;\;|A'|<0\;\color{red}\implies\;\text{Hyperbola}\\{}\\r(A)=3\,,\,\,r(A')=1\;\color{red}\implies\;\text{Parabola}\end{cases}$$

and etc. The signature kicks in for one pair of parallel lines ($\;\sigma(A)=0\;$ and the ranks one less than the maximal possible), and for empty set ($\;\sigma(A)=2\;$ and the ranks as before)

The above is related to bilinear forms and their close relatives, quadratic forms, Sylvester's Inertia Theorem and etc. It is a little advanced material in linear algebra and it is usually covered, as far as I know, in a second course in this subject at undergraduate level.