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Reducing the general equation of a line of the second order to the canonical view

2
2
where D = Ax 0 + By 0 + D; E = Bx 0 + Cy 0 + E; F = Ax + 2Bx 0 y 0 + Cy + 2Dx 0 +
′
′
′
0
0
2Ey 0 + F. In equation (7.31) coefficients D and E will be equal to zero, if coordinates of point
′
′
O (x 0 ; y 0 ) are chosen in order to be the solution of the following system of equations:
′
{
Ax 0 + By 0 + D = 0,
(7.32)
Bx 0 + Cy 0 + E = 0.
2
As AC − B ̸= 0, then this system has the only solution relatively to x 0 and y 0 , and equation
(7.31) will look like:
′ ′2
′ ′2
′
A x + 2B x y + C y + F = 0, (7.33)
′ ′ ′
Suppose, that a rectangular system of coordinates O x y is the result of turning system of coor-
′ ′′ ′′
dinates Ox y on angle α. Then the following formulas connect old coordinates x , y with new ones:
′
′
′ ′
′′
′ ′′ ′′
′′
′′
′
′
′′
x = x cos α − y sin α, y = x sin α + y cos α. So, in the system of coordinates O x y
equation (7.33) looks like:
′ ′′2
′ ′′′2
A x + 2B x y + C y + F = 0, (7.34)
′
′ ′′ ′′
2
2
2
2
whereA = A cos α+2B cos α sin α+C sin α,B = −A cos α sin α+B(cos α−sin α)+
′
′
2
2
C cos α sin α,C = A sin α−2B cos α sin α+C cos α.Havingchosenangleαtobecoefficient
′
′
B in equation (7.34) equal to zero, we will get the equation relatively to α : 2B cos 2α = (A −
C) sin 2α. If A = C, then cos 2α = 0, and α = π . If A ̸= C, then α = 1 arctg 2B , and
4 2 A−C
′2
′2
equation (7.34) looks like: Ax + Cy + F = 0. 2
Remark 7.4. Equation (7.32) is called the equation of the center of a line of the
second order, and point (x 0 ; y 0 ), which is found from the last system, is called the center
of the line. Therefore, in order to lead general equation (7.1) to its canonical view, the
following steps must be taken:
1. number of terms of the first degree should be the least;
2. to destroy a free term (if it possible);
3. to destroy a term with multiplication xy.
Example 7.1. Bring to its canonical view the equation:
2
2
17x + 12xy + 8y − 46x − 28y + 17 = 0.
′
Solution. First of all, perform parallel shift of axes following the formulas: x = x + x 0 ; y = y + y 0 .
′
Having substituted these coordinates in the left part of the given equation, we will have:
2
′2
2
′2
′
′ ′
17x + 12xy + 8y − 46x − 28y + 17 = 17x + 12x y + 8y + 2(17x 0 + 6y 0 − 23)x +
2
2
′
+2(6x 0 + 8y 0 − 14)y + 17x + 12x 0 y 0 + 8y − 46x 0 − 28y 0 + 17.
0 0
{ {
Let’s choose x 0 and y 0 as a solution of the system: 17x 0 + 6y 0 − 23= 0, ⇒ x 0 = 1,
6x 0 + 8y 0 − 14 = 0 y 0 = 1.
Calculate the free term:
2
2
F = 17x + 12x 0 y 0 + 8y − 46x 0 − 28y 0 + 17 = (17x 0 + 6y 0 − 23)x 0 +
′
0 0
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