Worked examples · Sheet 3

Reductions to canonical form with rotation and translation

Theory used: Rigid motion The matrix after a rigid motion Canonical form of central conics Reduction procedure Axes of symmetry Asymptotes of a hyperbola

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Find a rotation that brings the conic

5 x^{2} + 6 x y + 5 y^{2} - 8 = 0

to canonical form, the canonical form and the axes of symmetry.

A_{0} = (5335), p_{A_{0}} (λ) = λ^{2} - 10 λ + 16 = (λ - 2) (λ - 8)

V_{2} = ⟨(1, - 1)⟩, V_{8} = ⟨(1, 1)⟩, R = \frac{1}{2} (1 - 1 11), det R = 1

x = \frac{X + Y}{2}, y = \frac{- X + Y}{2}

5 (x^{2} + y^{2}) + 6 x y - 8 = 5 (X^{2} + Y^{2}) + 3 (Y^{2} - X^{2}) - 8 = 2 X^{2} + 8 Y^{2} - 8

\frac{X ^{2}}{4} + Y^{2} = 1

Ellipse with center at the origin, semi-axes $2$ and $1$ ; major axis $y = - x$ , minor axis $y = x$ .

The conic has no first-degree terms, so the center is the origin and a rotation is enough. The eigenvalues of $A_{0}$ are $2$ and $8$ . For $λ = 2$ the matrix $A_{0} - 2 I$ , with $I$ the identity matrix, has rows equal to $(3, 3)$ , so the eigenspace is given by $x + y = 0$ and is spanned by $(1, - 1)$ ; for $λ = 8$ the rows are $(- 3, 3)$ and $(3, - 3)$ , and the eigenspace is spanned by $(1, 1)$ . Normalizing the two eigenvectors gives the columns of $R$ , which has determinant $\frac{1 + 1}{2} = 1$ . In the substitution $x^{2} + y^{2} = X^{2} + Y^{2}$ , because the rotation preserves distances from the origin, and $x y = \frac{Y ^{2} - X ^{2}}{2}$ . The equation becomes $2 X^{2} + 8 Y^{2} = 8$ , that is, the ellipse $\frac{X ^{2}}{4} + Y^{2} = 1$ . The semi-axis $2$ lies on the $X$ -axis, which has the direction of $(1, - 1)$ , so the major axis is the line $y = - x$ and the minor axis is $y = x$ .

Exercise 2

Reduce to canonical form the conic

17 x^{2} + 12 x y + 8 y^{2} - 22 x + 4 y - 7 = 0,

writing the rigid motion and the equations of the axes.

A_{0} = (17668), I_{2} = 136 - 36 = 100, p_{A_{0}} (λ) = λ^{2} - 25 λ + 100 = (λ - 5) (λ - 20)

V_{5} = ⟨(1, - 2)⟩, V_{20} = ⟨(2, 1)⟩

{17 x + 6 y - 11 = 0 6 x + 8 y + 2 = 0 ⟹ (x_{0}, y_{0}) = (1, - 1), f (1, - 1) = 17 - 12 + 8 - 22 - 4 - 7 = - 20

(x y) = \frac{1}{5} (1 - 2 21) (X Y) + (1 - 1)

5 X^{2} + 20 Y^{2} - 20 = 0 ⟹ \frac{X ^{2}}{4} + Y^{2} = 1

major axis: 2 x + y - 1 = 0, minor axis: x - 2 y - 3 = 0

Ellipse with center $(1, - 1)$ and semi-axes $2$ and $1$ .

The ellipse

17 x^{2} + 12 x y + 8 y^{2} - 22 x + 4 y - 7 = 0

with the axes

2 x + y - 1 = 0

and

x - 2 y - 3 = 0

(dashed), which meet at the center

(1, - 1)

We compute the eigenvalues and eigenvectors of $A_{0}$ . For $λ = 5$ the matrix $A_{0} - 5 I$ has first row $(12, 6)$ , so the eigenspace is $2 x + y = 0$ , spanned by $(1, - 2)$ ; for $λ = 20$ the first row is $(- 3, 6)$ , the eigenspace is $x - 2 y = 0$ , spanned by $(2, 1)$ . The two vectors have norm $5$ , and the matrix with columns $\frac{( 1 , - 2 )}{5}$ and $\frac{( 2 , 1 )}{5}$ has determinant $\frac{1 + 4}{5} = 1$ , so it is a rotation. The center is found with the system of the center: from the second equation $x = \frac{- 8 y - 2}{6}$ , and the solution is $(1, - 1)$ , which is checked by substituting into both equations. The polynomial evaluated at the center equals $- 20$ , and by the canonical form of central conics the equation in the new coordinates is $5 X^{2} + 20 Y^{2} - 20 = 0$ . As a check, $I_{3} = I_{2} \cdot f (1, - 1) = - 2000$ , which coincides with $det A$ . The axes are the lines through the center with the directions of the eigenvectors: with direction $(1, - 2)$ we obtain $2 x + y - 1 = 0$ , which contains the semi-axis $2$ and is therefore the major axis; with direction $(2, 1)$ we obtain $x - 2 y - 3 = 0$ .

Exercise 3

Reduce to canonical form the conic

3 x^{2} - 8 x y - 3 y^{2} + 2 x + 14 y - 13 = 0

and write the equations of the axes and of the asymptotes.

A = 3 - 4 1 - 4 - 3 7 17 - 13, I_{2} = - 9 - 16 = - 25, I_{1} = 0

{3 x - 4 y + 1 = 0 - 4 x - 3 y + 7 = 0 ⟹ (x_{0}, y_{0}) = (1, 1), f (1, 1) = 3 - 8 - 3 + 2 + 14 - 13 = - 5

I_{3} = I_{2} \cdot f (1, 1) = 125, p_{A_{0}} (λ) = λ^{2} - 25 ⟹ λ_{1} = 5, λ_{2} = - 5

V_{5} = ⟨(2, - 1)⟩, V_{- 5} = ⟨(1, 2)⟩, (x y) = \frac{1}{5} (2 - 1 12) (X Y) + (11)

5 X^{2} - 5 Y^{2} - 5 = 0 ⟹ X^{2} - Y^{2} = 1

axes: x + 2 y - 3 = 0, 2 x - y - 1 = 0

f (x, y) - f (1, 1) = 3 x^{2} - 8 x y - 3 y^{2} + 2 x + 14 y - 8 = (3 x + y - 4) (x - 3 y + 2)

Rectangular hyperbola with center $(1, 1)$ , canonical form $X^{2} - Y^{2} = 1$ , asymptotes $3 x + y - 4 = 0$ and $x - 3 y + 2 = 0$ .

The rectangular hyperbola

3 x^{2} - 8 x y - 3 y^{2} + 2 x + 14 y - 13 = 0

with the asymptotes

3 x + y - 4 = 0

and

x - 3 y + 2 = 0

(dashed), perpendicular and intersecting at the center

(1, 1)

Since $I_{2} = - 25 < 0$ and $I_{1} = 0$ , the conic, if it is not degenerate, is a rectangular hyperbola. The center solves the system of the center: multiplying the first equation by $4$ and the second by $3$ and adding gives $- 25 y + 25 = 0$ , hence $y = 1$ and $x = 1$ . The value of the polynomial at the center is $- 5$ , hence $I_{3} = I_{2} \cdot (- 5) = 125 \neq = 0$ and the conic is indeed non-degenerate. The eigenvalues are $\pm 5$ ; the eigenspace of $5$ is $x + 2 y = 0$ , spanned by $(2, - 1)$ , that of $- 5$ is $2 x - y = 0$ , spanned by $(1, 2)$ , and the matrix with these normalized columns has determinant $\frac{4 + 1}{5} = 1$ . The canonical form is $5 X^{2} - 5 Y^{2} = 5$ , that is, $X^{2} - Y^{2} = 1$ . The axes are the lines through $(1, 1)$ with directions $(2, - 1)$ and $(1, 2)$ ; the first, $x + 2 y - 3 = 0$ , is the transverse axis. For the asymptotes we use the rule $f (x, y) - f (x_{0}, y_{0}) = 0$ : the polynomial $3 x^{2} - 8 x y - 3 y^{2} + 2 x + 14 y - 8$ factors as the product $(3 x + y - 4) (x - 3 y + 2)$ , as can be checked by expanding. The two asymptotes have normal vectors $(3, 1)$ and $(1, - 3)$ , which are orthogonal, in agreement with the fact that the hyperbola is rectangular.