Canonical Forms
Canonical Forms of Second Order Curves
Second order curves can be described by the following canonical equations:
| Real Ellipse |
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) |
| Imaginary Ellipse |
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = -1\) |
| Hyperbola |
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) |
| Parabola |
\(y^2 = 2px\) |
| Pair of Imaginary Intersecting Lines (a Real Point) |
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 0\) |
| Pair of Real Intersecting Lines |
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0\) |
| Pair of Real Parallel Lines |
\(x^2 - d^2 = 0\) |
| Pair of Coincident Lines |
\(x^2 = 0\) |
| Pair of Imaginary Parallel Lines |
\(x^2 + d^2 = 0\) |
Reduction to Canonical Forms
A general equation of a second-order curve in Cartesian coordinates is given by
\[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]
The general second-order equation can be transformed into canonical form by changing the coordinate system. Such a change may include both parallel translation and rotation of the coordinate system.
Case 1. The General Equation Does Not Have An XY-Term
If the general equation does not contain an xy-term, that is, if the coefficient \(B = 0,\) then such an equation can be reduced into canonical form using only a parallel transfer of the coordinate system. Algebraic transformations are based on completing squares for terms with variables x and y.
Example
Write the equation of the curve in canonical form:
\[9x^2 + 4y^2 + 6x - 4y - 2 = 0.\]
Solution.
Let's group terms with x and y:
\[\left({9x^2 + 6x}\right) + \left({4y^2 - 4y}\right) - 2 = 0, \Rightarrow 9\left({x^2 + \frac{2}{3}x}\right) + 4\left({y^2 - y}\right) - 2 = 0.\]
Complete the square for x2 and y2 terms:
\[9\left({x^2 + \frac{2}{3}x \pm\frac{1}{9}}\right) + 4\left({y^2 - y \pm\frac{1}{4}}\right) - 2 = 0,\]
\[9\left({x + \frac{1}{3}}\right)^2 - 1 + 4\left({y - \frac{1}{2}}\right)^2 -1 - 2 = 0,\]
\[9\left({x + \frac{1}{3}}\right)^2 + 4\left({y - \frac{1}{2}}\right)^2 = 4.\]
Divide the left and right sides of the equation by 4:
\[\frac{9\left({x + \frac{1}{3}}\right)^2}{4} + \frac{\left({y - \frac{1}{2}}\right)^2}{1} = 1, \Rightarrow \frac{\left({x + \frac{1}{3}}\right)^2}{\frac{4}{9}} + \frac{\left({y - \frac{1}{2}}\right)^2}{1} = 1.\]
Recall that the canonical equation of an ellipse centered at \(\left({x_0,y_0}\right)\) is given by
\[\frac{\left({x - x_0}\right)^2}{a^2} + \frac{\left({y - y_0}\right)^2}{b^2} = 1\]
Thus, the canonical equation obtained above is the equation of an ellipse centered at point \(\left({-\frac{1}{3},\frac{1}{2}}\right)\) with the semi-major and semi-minor axes \(a = \frac{2}{3}\) and \(b = 1,\) respectively.
Case 2. The General Equation Has An XY-Term
If the general equation contains a Bxy term, then you first need to get rid of it by rotating the coordinate system by a certain angle φ, which is defined by the formula
\[\cot2\varphi = \frac{A-C}{B}\]
where it is assumed that the curve is given by the general equation.
In the special case when \(A = C,\) the rotation angle is \(45^\circ:\)
\[A = C, \Rightarrow \cot2\varphi = \frac{A-C}{B} = \frac{0}{B} = 0, \Rightarrow 2\varphi = \pm 90^\circ, \Rightarrow \varphi = \pm 45^\circ.\]
After rotation by angle φ we get a new XY-coordinate system in which the equation of the curve does not contain the XY-term, that is, Case 2 reduces to Case 1 considered above.
Example
Write the equation of the curve in canonical form:
\[x^2 - 2xy + y^2 + 2x + 2y = 0.\]
Solution.
Identify the coefficients in the general equation and calculate the angle of rotation:
\[A = 1, B = -2, C = 1.\]
Since \(A = C,\) then we choose the angle of rotation \(\varphi = 45^\circ = \frac{\pi}{4}.\)
Let's denote the new coordinates after rotation as X and Y. The old coordinates are expressed in terms of the new ones using relations
\[\left\{ \begin{array}{l}
x = X\cos\varphi - Y\sin\varphi\\
y = X\sin\varphi + Y\cos\varphi
\end{array} \right..\]
If \(\varphi = \frac{\pi}{4}\) then we get
\[\sin\varphi = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2},\;\cos\varphi = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}.\]
Therefore, the coordinate transformation is described by the equations
\[\left\{ \begin{array}{l}
x = X\frac{\sqrt{2}}{2} - Y\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\left({X - Y}\right)\\
y = X\frac{\sqrt{2}}{2} + Y\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\left({X + Y}\right)
\end{array} \right..\]
Substitute x and y into the equation of the curve:
\[\left({\frac{\sqrt{2}}{2}}\right)^2\left({X - Y}\right)^2 - 2\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}\left({X - Y}\right)\left({X + Y}\right) + \left({\frac{\sqrt{2}}{2}}\right)^2\left({X + Y}\right)^2 + 2\cdot\frac{\sqrt{2}}{2}\left({X - Y}\right) + 2\cdot\frac{\sqrt{2}}{2}\left({X + Y}\right) = 0,\]
\[\frac{1}{2}\left({X^2 - 2XY + Y^2}\right) - \left({X^2 - Y^2}\right) + \frac{1}{2}\left({X^2 + 2XY + Y^2}\right) + \sqrt{2}\left({X - Y}\right) + \sqrt{2}\left({X + Y}\right) = 0,\]
\[\cancel{\frac{1}{2}X^2} - \cancel{XY} + \frac{1}{2}Y^2 - \cancel{X^2} + Y^2 + \cancel{\frac{1}{2}X^2} + \cancel{XY} + \frac{1}{2}Y^2 + \sqrt{2}X - \cancel{\sqrt{2}Y} + \sqrt{2}X + \cancel{\sqrt{2}Y} = 0,\]
\[2Y^2 + 2\sqrt{2}X = 0.\]
This equation can be written as
\[X = -\frac{\sqrt{2}}{2}Y^2\]
In the new XY-coordinate system, we obtained the canonical equation of a parabola.
Solved Problems
Example 1.
Reduce the second-order equation to canonical form and identify the curve type:
\[4x^2 - y^2 - 16x + 6y + 23 = 0.\]
Solution.
Let's group the terms with x and y:
\[\left({4x^2 - 16x}\right) - \left({y^2 - 6y}\right) + 23 = 0.\]
Complete the squares by adding and subtracting the integers:
\[4\left({x^2 - 4x \pm 4}\right) - \left({y^2 - 6y \pm 9}\right) + 23 = 0,\]
\[4\left({x - 2}\right)^2 - 16 - \left({y - 3}\right)^2 + 9 + 23 = 0,\]
\[\left({y - 3}\right)^2 - 4\left({x - 2}\right)^2 = 16.\]
Divide both sides of the equation by 16:
\[\frac{\left({y - 3}\right)^2}{16} - \frac{\left({x - 2}\right)^2}{4} = 1.\]
As you can see, we have obtained the equation of a hyperbola. The center of the hyperbola is at the point \(\left({2,3}\right).\) Also this hyperbola has a vertical orientation, that is, its transverse axis is directed along the y-axis.
Example 2.
Reduce the second-order equation to canonical form and determine the type of the line:
\[9x^2 - 4y^2 + 36x - 16y + 20 = 0.\]
Solution.
Rewrite the equation in the form
\[\left({9x^2 + 36x}\right) - \left({4y^2 + 16y}\right) + 20 = 0.\]
Add and subtract the new integers in brackets to complete the squares:
\[9\left({x^2 + 4x \pm4}\right) - 4\left({y^2 + 4y \pm4}\right) + 20 = 0,\]
\[9\left({x + 2}\right)^2 - \cancel{36} - 4\left({y + 2}\right)^2 + \cancel{16} + \cancel{20} = 0,\]
so we get
\[9\left({x + 2}\right)^2 = 4\left({y + 2}\right)^2.\]
The canonical equation is written as
\[\frac{\left({x + 2}\right)^2}{4} - \frac{\left({y + 2}\right)^2}{9} = 0.\]
Let's go back to the previous equation and take the square root of both sides:
\[3\left({x + 2}\right) = \pm 2\left({y + 2}\right).\]
We are faced with two solutions here. In the first case we get the following equation:
- \(3x + 6 = 2y + 4,\;\) or \(\;3x - 2y + 2 = 0.\)
The second solution has the form
- \(3x + 6 = -2y -4,\;\) or \(\;3x + 2y + 10 = 0.\)
Therefore, the original equation describes two real intersecting straight lines.
Example 3.
Determine the type of second-order curve and write its canonical equation:
\[x^2 + x - 6 = 0.\]
Solution.
Complete the square in this equation:
\[x^2 + x - 6 = 0,\;\Rightarrow x^2 + x + \frac{1}{4} - \frac{25}{4} = 0,\;\Rightarrow \left({x+\frac{1}{2}}\right)^2 - \left({\frac{5}{2}}\right)^2 = 0.\]
We have obtained a canonical equation that describes a pair of parallel lines. Let's define the equations of these two straight lines:
\[\left({x+\frac{1}{2}}\right)^2 - \left({\frac{5}{2}}\right)^2 = 0,\;\Rightarrow x + \frac{1}{2} = \pm\frac{5}{2}.\]
- \(x + \frac{1}{2} = \frac{5}{2},\;\Rightarrow x = 2;\)
- \(x + \frac{1}{2} = -\frac{5}{2},\;\Rightarrow x = -3.\)
Example 4.
Reduce the second-order equation to canonical form and identify the type of the curve:
\[5x^2 + 4xy + 8y^2 - 32x - 56y + 80 = 0.\]
Solution.
It is easy to see that the given general equation has the following leading coefficients:
\[A = 5,\; B = 4,\; C = 8.\]
Since the equation contains an xy-term, we will first rotate the coordinate system to get rid of this term. The angle of rotation φ is determined by the relation
\[\cot2\varphi = \frac{A - C}{B} = \frac{5-8}{4} = -\frac{3}{4}.\]
The cotangent is negative, therefore the angle φ lies in the second quadrant:
\[\frac{\pi}{2} \lt 2\varphi \lt \pi.\]
In that case we have
\[\frac{\pi}{4} \lt \varphi \lt \frac{\pi}{4},\;\Rightarrow \sin\varphi \gt 0,\;\cos\varphi \gt 0,\;\cos2\varphi \lt 0.\]
Now we will use some trigonometric identities to construct the new XY-coordinate system. We need the formula
\[\frac{1}{\sin^2 2\varphi} = \cot^2 2\varphi + 1, \;\Rightarrow \sin^2 2\varphi = \frac{1}{\cot^2 2\varphi + 1}.\]
Substitute the cotangent value:
\[\sin^2 2\varphi = \frac{1}{\cot^2 2\varphi + 1} = \frac{1}{\left({-\frac{3}{4}}\right)^2 + 1} = \frac{1}{\frac{9}{16}+1} = \frac{1}{\frac{25}{16}} = \frac{16}{25}.\]
Find the cosine of the angle 2φ:
\[\cos^2 2\varphi = 1 - \sin^2 2\varphi = 1 - \frac{16}{25} = \frac{9}{25}.\]
From here we get \(\cos 2\varphi = -\frac{3}{5}.\) Now we can determine \(\sin \varphi\) and \(\cos \varphi:\)
\[\sin \varphi = \sqrt{\frac{1-\cos2\varphi}{2}} = \sqrt{\frac{1-\left({-\frac{3}{5}}\right)}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}};\]
\[\cos \varphi = \sqrt{\frac{1+\cos2\varphi}{2}} = \sqrt{\frac{1+\left({-\frac{3}{5}}\right)}{2}} = \sqrt{\frac{2}{10}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}.\]
After rotating the coordinate system by the angle φ, the old xy-coordinates are expressed in terms of the new XY-coordinates as follows:
\[\left\{ \begin{array}{l}
x = X\cos\varphi - Y\sin\varphi\\
y = X\sin\varphi + Y\cos\varphi
\end{array} \right.,\;\Rightarrow \left\{ \begin{array}{l}
x = X\frac{1}{\sqrt{5}} - Y\frac{2}{\sqrt{5}} = \frac{1}{\sqrt{5}}\left({X - 2Y}\right)\\
y = X\frac{2}{\sqrt{5}} + Y\frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}}\left({2X + Y}\right)
\end{array} \right..\]
We substitute these expressions for x and y into the equation of the curve:
\[5\cdot\left(\frac{1}{\sqrt{5}}\right)^2\left({X - 2Y}\right)^2 + 4\cdot\frac{1}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}}\left({X - 2Y}\right)\left({2X + Y}\right) + 8\cdot\left(\frac{1}{\sqrt{5}}\right)^2\left({2X + Y}\right)^2 - 32\cdot\frac{1}{\sqrt{5}}\left({X - 2Y}\right) - 56\cdot\frac{1}{\sqrt{5}}\left({2X + Y}\right) + 80 = 0,\]
\[X^2 - 4XY + 4Y^2 + \frac{4}{5}\left({2X^2 - 4XY + XY - 2Y^2}\right) + \frac{8}{5}\left({4X^2 + 4XY + Y^2}\right) - \frac{32}{\sqrt{5}}\left({X-2Y}\right) - \frac{56}{\sqrt{5}}\left({2X+Y}\right) + 80 = 0,\]
\[X^2 - \cancel{4XY} + 4Y^2 + \frac{8}{5}X^2 - \cancel{\frac{12}{5}XY} - \frac{8}{5}Y^2 + \frac{32}{5}X^2 + \cancel{\frac{32}{5}XY} + \frac{8}{5}Y^2 - \frac{32}{\sqrt{5}}X + \frac{64}{\sqrt{5}}Y - \frac{112}{\sqrt{5}}X - \frac{56}{\sqrt{5}}Y\ + 80 = 0.\]
Combining the like terms we have
\[\frac{45}{5}X^2 + \frac{20}{5}Y^2 - \frac{144}{\sqrt{5}}X + \frac{8}{\sqrt{5}}Y + 80 = 0,\]
or
\[9X^2 + 4Y^2 - \frac{144}{\sqrt{5}}X + \frac{8}{\sqrt{5}}Y + 80 = 0.\]
As you can see, the equation of the curve does not contain an XY-term in the new coordinate system. So now it is enough to complete squares to convert the equation into canonical form and determine the curve type.
\[\left({9X^2 - \frac{144}{\sqrt{5}}X}\right) + \left({4Y^2 + \frac{8}{\sqrt{5}}Y}\right) + 80 = 0,\]
\[9\left[{X^2 - \frac{16}{\sqrt{5}}X \pm\left({\frac{8}{\sqrt{5}}}\right)^2}\right] + 4\left[{Y^2 + \frac{2}{\sqrt{5}}Y \pm\left({\frac{1}{\sqrt{5}}}\right)^2}\right] + 80 = 0,\]
\[9\left({X - \frac{8}{\sqrt{5}}}\right)^2 - \frac{576}{5} + 4\left({Y + \frac{1}{\sqrt{5}}}\right)^2 - \frac{4}{5} + 80 = 0,\]
\[9\left({X - \frac{8}{\sqrt{5}}}\right)^2 + 4\left({Y + \frac{1}{\sqrt{5}}}\right)^2 = 36.\]
Divide both sides of the equation by 36 and get the canonical equation of an ellipse in the new XY-coordinate system:
\[\frac{\left({X - \frac{8}{\sqrt{5}}}\right)^2}{4} + \frac{\left({Y + \frac{1}{\sqrt{5}}}\right)^2}{9} = 1.\]
