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0EXAMPLES OF CANONIZATION AND DETERMINATION OF THE TYPE OF SECOND ORDER CURVES AND THEIR DIFFERENT SOLUTIONS
Abdullaeva D.M.
Uzbekistan-Finland Pedagogical Institute, Student of 307-group of "Mathematics and Computer
Science" specialty https://doi.org/10.5281/zenodo.13763466
Abstract. The given article presents different recommendations and methods for canonization of quadratic curves, as well as examples of their solution.
Keywords: second-order curve, canonical representation, ellipse, hyperbola, parabola, intersecting lines, parallel lines, invariant, simplification.
Simplification of general equations
a
li
J y
x~ + 2a12xy + a12y~ + + 2a23 y + Ö33 =0(1)
It is important to suggest that the given study is concerned with determining the type of the second-order line given by the general equation (1) and making it canonic. We will also first get acquainted with the second-order line itself.
Definition: the lines defined by quadratic equations with respect to the variables x and y are called quadratic lines. Second-order lines play an important role in architecture, astronomy, mechanics and other branches of science and technology. Second-order lines can be divided into three groups:
1) Second-order lines with one center of symmetry
2) This group includes second-order lines without a center of symmetry.
3) The third group includes lines whose center of symmetry is a straight line. x2 + y2 = r2(0: 0), The second-order line given by the equation is a circle of radius r with center at the
point (0:0), ^2 + ^2 = 1 an ellipse whose center is at the origin, — = 1 hyperbola
given by the equation, y2 = 2px defines a parabola.
Let a general equation of the second degree with two variables be given:
«j j x" + 2a[2xy + a->2y~ + 2a^X + 2a23 y + Ujj =0 (!)
In the above equation, a11 and a22 cannot be equal to zero simultaneously. It is clear from the given information that the canonical equations of the second-order curves considered above are eigenvalues of equation (1).
If aii=1, ai2=0, a22=1, ai3=0, a23=0, a33=-r2 (1) equiation x2 + y2 = r2 takes the form, that is, this equation defines a circle.
1 1 X2 V2
a11 = —, a12 = 0, a22 a13=0, a23=0, a33=-1 (1) equiation — + tz=1 defines the
ellipse.
1 1
In a11 a12 = 0, a22 = ai3=0, a23=0, a33=-1 (1) equiation
x2 y2
— —- = 1 defines the hyberbole.
a2 b2 J
an = 0, a12 = 0, a22 = 1, ai3=-p, a23=0, a33=0 (1) equiation y2 = 2px will have the form, and therefore the equation of a parabola.
The following suggestion can be drawn from the above substitutions: therefore, an arbitrary second-order curve, given in general form, can be reduced to a canonical form. When we mean reducing a second-order line, given by a general equation, to a canonical state. If in the system of equations (1) a second-order curve is given in the coordinate system R, then, by rotating the coordinate system by an angle, we can go to a rectangular coordinate system R. If there is a12 ^ 0 in the equation (1), then we construct the characteristic equation of the line I2 — (a11 + a22)l + a11a22 — q22 and find its roots A1,A2 according to the formula sin a = -j===== , cosa =
i 1 0 , calculate tg a = ai1. Taking into account the above, the coordinate vectors i', / of the
standard (o, /,_/') are determined, created by rotating the coordinate system by an angle a. i =icosa + /sina / = y'cosa — isina Line equation in new rapper looks like this
l1x 2 + l2y 2 + 2a13x + 2a23y + a33 = 0 and the coefficients a13, a23 are determined by the following formulas:
a13 = a13cosa + a23sina, a23 = a23cosa — a13sina. Taking into consideration all of these let us get acquainted with the next method of canonization of second-order strings, this method is called the "invariant" method.
/1 = an + «12 (1)
U =
K. =
(2) (3)
a11 a12 a21 a22 a11 a12 a13 a21 a22 a23 a31 a32 a33
Definitely, the expressions we mentioned above are called invariants of lines of the second order with respect to the change of a rectangular coordinate system to another coordinate system, and the expressions a12 = a21 a31 = a13 a32 = a23 are always appropriate here.
Let us determine the type and canonical equation of the second-order line according to the above invariants. We will give the characteristic equation for the specified invariants and determine the type of line according to the equation.
I2 — IjA + /2 = 0, the roots A1 and A2 of this equation consist of real numbers. The lines of the second order can be divided into three groups, as we said above. The first group includes lines with a single center of symmetry, an ellipse, an abstract ellipse, two intersecting abstract lines, a hyperbole, and two intersecting real lines. /2 ^ 0 necessary and sufficient for a line of the second order to have a single center.
The second group includes lines that do not have a center of symmetry, i.e. a parabole. For a line to be a parabole, it is necessary and sufficient that /2 = 0, ^ 0 . The third group includes lines whose centers of symmetry form a straight line, two parallel lines, or two overlapping lines. In order for the centers of symmetry of a second-order line to form a straight line, it is necessary and sufficient that the condition /2 = 0, = 0 can be met. The equation of the lines of the first
group, formed by changing the rectangular coordinate system, is l1x2 + l2y2 + —
b
= 0.
Moreover, the equation of the second group: I1y2 ±2J-^3x=0. We can reduce the
equation of the third group to the form I1x2 + — = 0 . The necessary and sufficient conditions for
h
the membership of second-order lines in the specified groups are expressed by the relations between their invariants as follows:
I. It is an ellipse if the conditions l2 > 0, I1, K3 < 00 are met, an abstract ellipse is l2 > 0, l1,K3 > 0. Two intersecting abstract lines I2 > 0, K3 = 0
Hyperbole I2 < 0, K3 ^ 0
II. Parabole l2 = 0, K3^0
III. Two parallel lines l2 = 0, K3 = 0, K2 > 0
The straight line is blocked I2 = 0, K3 = 0, K2 = 0. Now let us make the following example canonical by the method of invariants;
x2 - 4xy + 4y2 -6x + 8y - 6 = 0 Solution: a11 = 1,a22 = 4,a12 = a21 = -2,a13 = a31 = -3,
a23 = a23 = 4, a,33 -6, Ii = an + ai2 = 1 + 4 = 5
h =
a11 a12 _ 1 -2
a21 a22 -2 4
a11 a12 a13 1 -2 -3
K3 = a21 a22 a23 = -2 4 4
a31 a32 a33 -3 4 -6
=4-4=0
= -24 + 24 + 24- 36 + 24- 16 =-4
By summerising it can be suggested that since I2 = 0, K3 ^ 0 the equation of a parabole, we will now bring this equation to the canonical state.
I1y2 ± 2 l-y-x = 0; ^5y2 ±2 J—= 0
i 4 i 4
5y2 — -j=x = 0 ^ y2 = x has the same canonical form as the equation above.
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