Worked examples · Sheet 2

Canonical forms of central conics

Theory used: Canonical form of central conics Center of a conic Classification of non-degenerate conics Orthogonal invariants Axes of symmetry

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Find the center and the canonical form of the conic

x^{2} + 4 y^{2} - 2 x + 16 y + 13 = 0.

x^{2} - 2 x + 4 y^{2} + 16 y + 13 = (x - 1)^{2} - 1 + 4 (y + 2)^{2} - 16 + 13 = (x - 1)^{2} + 4 (y + 2)^{2} - 4

X = x - 1, Y = y + 2 ⟹ X^{2} + 4 Y^{2} = 4 ⟹ \frac{X ^{2}}{4} + Y^{2} = 1

I_{3} = - 16, I_{2} = 4, \frac{I _{3}}{I _{2}} = - 4

Ellipse with center $(1, - 2)$ and semi-axes $2$ and $1$ , with canonical form $\frac{X ^{2}}{4} + Y^{2} = 1$ .

The equation has no $x y$ term, so $A_{0}$ is already diagonal and no rotation is needed. We complete the squares: $x^{2} - 2 x = (x - 1)^{2} - 1$ and $4 y^{2} + 16 y = 4 (y + 2)^{2} - 16$ . Adding the constant terms, the equation becomes $(x - 1)^{2} + 4 (y + 2)^{2} = 4$ , and the translation that moves the origin to $(1, - 2)$ gives $X^{2} + 4 Y^{2} = 4$ . Dividing by $4$ gives the ellipse with semi-axis $2$ along the $X$ -axis and semi-axis $1$ along the $Y$ -axis. The result agrees with the canonical form of central conics: the eigenvalues of $A_{0}$ are $1$ and $4$ and $\frac{I _{3}}{I _{2}} = - 4$ , so the canonical form is $X^{2} + 4 Y^{2} - 4 = 0$ .

Exercise 2

Find the center, the canonical form and the semi-axes of the conic

2 x^{2} + 4 x y + 5 y^{2} - 6 y - 3 = 0.

A = 220 25 - 3 0 - 3 - 3, I_{3} = 2 \cdot (- 15 - 9) - 2 \cdot (- 6 - 0) = - 36, I_{2} = 6, I_{1} = 7

p_{A_{0}} (λ) = λ^{2} - 7 λ + 6 = (λ - 1) (λ - 6) ⟹ λ_{1} = 1, λ_{2} = 6

{2 x + 2 y = 0 2 x + 5 y - 3 = 0 ⟹ (x_{0}, y_{0}) = (- 1, 1)

X^{2} + 6 Y^{2} - 6 = 0 ⟹ \frac{X ^{2}}{6} + Y^{2} = 1

V_{1} = ⟨(2, - 1)⟩, V_{6} = ⟨(1, 2)⟩

Real ellipse with center $(- 1, 1)$ , semi-axes $6$ and $1$ ; the major axis is the line $x + 2 y - 1 = 0$ .

The ellipse

2 x^{2} + 4 x y + 5 y^{2} - 6 y - 3 = 0

, with center

(- 1, 1)

, and the major axis

x + 2 y - 1 = 0

(dashed)

Since $I_{3} = - 36 \neq = 0$ and $I_{2} = 6 > 0$ with $I_{1} I_{3} < 0$ , the conic is a real ellipse. The eigenvalues of $A_{0}$ are the roots of the characteristic polynomial, that is, $1$ and $6$ . The center solves the system of the center: from the first equation $x = - y$ , and substituting into the second gives $3 y = 3$ , hence $y = 1$ and $x = - 1$ . By the canonical form of central conics the constant term is $\frac{I _{3}}{I _{2}} = - 6$ , and the canonical equation is $X^{2} + 6 Y^{2} = 6$ ; dividing by $6$ the semi-axes $6$ and $1$ can be read off. The semi-axis $6$ corresponds to the eigenvalue $1$ , so the major axis is the line through $(- 1, 1)$ with direction $(2, - 1)$ , that is, $x + 2 y - 1 = 0$ . As a check, the polynomial evaluated at the center equals $2 - 4 + 5 - 6 - 3 = - 6$ , equal to $\frac{I _{3}}{I _{2}}$ .

Exercise 3

Find the center and the canonical form of the conic

x^{2} + 6 x y + y^{2} + 4 x - 4 y - 8 = 0.

A = 132 31 - 2 2 - 2 - 8

I_{3} = 1 \cdot (- 8 - 4) - 3 \cdot (- 24 + 4) + 2 \cdot (- 6 - 2) = - 12 + 60 - 16 = 32

I_{2} = 1 - 9 = - 8, p_{A_{0}} (λ) = λ^{2} - 2 λ - 8 = (λ - 4) (λ + 2)

{x + 3 y + 2 = 0 3 x + y - 2 = 0 ⟹ (x_{0}, y_{0}) = (1, - 1), \frac{I _{3}}{I _{2}} = - 4

4 X^{2} - 2 Y^{2} - 4 = 0 ⟹ X^{2} - \frac{Y ^{2}}{2} = 1

Hyperbola with center $(1, - 1)$ , with canonical form $X^{2} - \frac{Y ^{2}}{2} = 1$ .

The determinant of $A$ equals $32$ and $I_{2} = - 8 < 0$ , so the conic is a hyperbola. The eigenvalues of $A_{0}$ are $4$ and $- 2$ , of opposite signs as they must be. The center is obtained from the second equation of the system, $y = 2 - 3 x$ , which substituted into the first gives $x + 6 - 9 x + 2 = 0$ , that is, $x = 1$ and hence $y = - 1$ . The constant term of the canonical form is $\frac{I _{3}}{I _{2}} = \frac{32}{- 8} = - 4$ ; dividing $4 X^{2} - 2 Y^{2} = 4$ by $4$ gives $X^{2} - \frac{Y ^{2}}{2} = 1$ , a hyperbola with semi-axes $a = 1$ and $b = 2$ . Since $a \neq = b$ , in agreement with $I_{1} = 2 \neq = 0$ , the hyperbola is not rectangular.