Euclidean invariants of the conics

From now on, we will suppose that $\overset{―}{a}$ and $A$ are the projection matrix and an infinite matrix, with reference to rectangular coordinates $(x, y)$ , in the equation of the conic.

Relative and absolute invariants

$\begin{matrix} D_{3} = d e t \overset{―}{A} \\ D_{2} = a c + a f + c f - (b^{2} + d^{2} + e^{2}) \\ d_{2} = d e t A = a c - b^{2} \\ d_{1} = T r A = a + c \end{matrix}$ These values are known as euclidean invariants.

The determination of the types of conics from the invariants can be obtained by means of the following table:

{\begin{cases} D_{3} \neq 0 {\begin{matrix} d_{2} > 0 ellipse {\begin{matrix} D_{3} d_{1} < 0 real \\ D_{3} d_{1} > 0 imaginay \end{matrix} \\ d_{2} < 0 hyperbola \\ d_{2} = 0 parabola \end{matrix} \\ D_{3} = 0 {\begin{matrix} d_{2} > 0 pair of imaginary conjugate lines \\ d_{2} < 0 pair of real lines \\ d_{2} = 0 {\begin{array}{c} D_{2} < 0 parallel real lines \\ D_{2} > 0 pair of imaginary parallel conjugate lines \\ D_{2} = 0 pair of coincidental lines \end{array} \end{matrix} \end{cases}

Now we are going to outline a few applications of the euclidean invariants:

Obtaining reduced equations: The reduced equation of the conic of the centered type is: $λ_{1} x^{2} + λ_{2} y^{2} + \frac{D_{3}}{d_{2}} = 0$ The canonical equation of a parabola is: $x^{2} + 2 \sqrt{- \frac{D_{3}}{d_{1}^{3}}} y = 0$ Finally, the reduced equation of two parallel straight lines is: $x^{2} + \frac{D_{2}}{d_{1}^{2}}$
Area of the ellipse: The area of the ellipse can be calculated using the formula: $A = π \sqrt{\frac{D_{3}^{2}}{d_{2}^{3}}}$
Angle of the asymptotes of a hyperbola: It is possible to determine the angle between the asymptotes of a hyperbola, or a pair of straight lines, by means of the formula $\cos^{2} α = \frac{d_{1}^{2}}{d_{1}^{2} - 4 d_{2}}$

Example

Classify the following conic and find its area: $x^{2} + 4 y^{2} + 4 x - 6 y + 9 = 0$

The associated matrix is $\overset{―}{A} = [\begin{matrix} 1 & 0 & 2 \\ 0 & 4 & - 3 \\ 2 & - 3 & 9 \end{matrix}]$ The associated euclidean invariants are: $D_{3} = d e t \overset{―}{A} = 36 - 16 - 9 = 11 d_{2} = 4 d_{1} = 1 + 4 = 5$ Notice that it is not necessary to compute $D_{2}$ since the determinant of the matrix is other than zero.

Therefore, following the classification algorithm, we conclude that this conic is an imaginary ellipse.

Finally, we can calculate its area by means of a formula that uses the Euclidean invariants: $A r e a = π \sqrt{\frac{D_{3}^{2}}{d_{2}^{3}}} = π \sqrt{\frac{121}{64}}$

Example

Classify the following conic using the euclidean invariants $q (x, y) = 3 x^{2} + 3 y^{2} - 6 x y + 4 y - 8 = 0$ The associated matrix is $\overset{―}{A} = [\begin{matrix} 3 & - 3 & - 3 \\ - 3 & 3 & 2 \\ - 3 & 2 & - 8 \end{matrix}]$ The associated euclidean invariants are: $D_{3} = d e t \overset{―}{A} = - 3 d_{2} = 0 d_{1} = 6$

Following the classification scheme of the euclidean invariants, we conclude that this is a parabola.