# 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

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:

Solution.

Let's group terms with *x* and *y*:

Complete the square for *x*^{2} and *y*^{2} terms:

Divide the left and right sides of the equation by 4:

Recall that the canonical equation of an ellipse centered at \(\left({x_0,y_0}\right)\) is given by

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

where it is assumed that the curve is given by the general equation.

In the special case when \(A = C,\) the rotation angle is \(45^\circ:\)

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:

Solution.

Identify the coefficients in the general equation and calculate the angle of rotation:

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

If \(\varphi = \frac{\pi}{4}\) then we get

Therefore, the coordinate transformation is described by the equations

Substitute *x* and *y* into the equation of the curve:

This equation can be written as

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*:

Complete the squares by adding and subtracting the integers:

Divide both sides of the equation by 16:

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

Add and subtract the new integers in brackets to complete the squares:

so we get

The canonical equation is written as

Let's go back to the previous equation and take the square root of both sides:

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:

We have obtained a canonical equation that describes a pair of parallel lines. Let's define the equations of these two straight lines:

- \(x + \frac{1}{2} = \frac{5}{2},\;\Rightarrow x = 2;\)
- \(x + \frac{1}{2} = -\frac{5}{2},\;\Rightarrow x = -3.\)

### 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:

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

The cotangent is negative, therefore the angle *φ* lies in the second quadrant:

In that case we have

Now we will use some trigonometric identities to construct the new *XY*-coordinate system. We need the formula

Substitute the cotangent value:

Find the cosine of the angle 2*φ*:

From here we get \(\cos 2\varphi = -\frac{3}{5}.\) Now we can determine \(\sin \varphi\) and \(\cos \varphi:\)

After rotating the coordinate system by the angle *φ*, the old *xy*-coordinates are expressed in terms of the new *XY*-coordinates as follows:

We substitute these expressions for *x* and *y* into the equation of the curve:

Combining the like terms we have

or

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.

Divide both sides of the equation by 36 and get the canonical equation of an ellipse in the new *XY*-coordinate system: