Conic Sections, Rotation of Axes

Rotation of Axes

In a rotation of axes, the origin remains fixed while the x-axis and the y-axis are rotated through an angle θ to a new position; the new positions of the x and y axes are denoted by x' and y', respectively as shown in figure.

y y y' α r P x' x O θ x' x
fig 6.4-2

Here, the point P has the coordinates (x, y) relative to the xy-plane. We can express x and y in terms of x', y' and θ.

As in figure, r denotes the distance from the origin to the point P, and α denotes the angle between the positive x'-axis and the ray from O through P.
Then, by using definition of sine and cosine, we have
x' = r cos α, y' = r sin α,
x = r cos(θ + α), y = r sin(θ + α).
Next,
x = r cos(θ + α)
= r(cos θ cos α - sin θ sin α)
= (r cos α)(cos θ) - (r sin α)(sin θ)
= x' cos θ - y' sin θ

Similarly,
y = r sin(θ + α)
= r(sin θ cos α + cos θ sin α)
= (r cos α)(sin θ) + (r sin α)(cos θ)
= x' sin θ + y' cos θ

Now we have the following theorem.

Theorem

If the x and y axes are rotated through an angle θ, then the coordinates (x, y) of a point P relatively to the xy-plane and the coordinates (x' , y') of the same point relative to the new x' and y' axes are related by the formula
x' = x cos θ + y sin θ
y' = -x sin θ + y cos θ.

Proof:
We have the rotation equations:
x = x' cos θ - y' sin θ
y = x' sin θ + y' cos θ.

Multiplying with sin θ and cos θ to the respective rotation equations gives:
x sin θ = x' sin θ cos θ - y' sin² θ (1)
y cos θ = x' sin θ cos θ + y' cos² θ (2)

Solving equations (1) and (2)
y' = -x sin θ + y cos θ

Multiplying with cos θ and sin θ to the respective rotation equations gives:
x cos θ = x' cos² θ - y' sin θ cos θ (3)
y sin θ = x' sin² θ + y' sin θ cos θ (4)
Adding equations (3) and (4) gives
x' = x cos θ + y sin θ
So, we have the formulas
x' = x cos θ + y sin θ
y' = -x sin θ + y cos θ.

Example 12
Find the new coordinates of the point (1, 2) if the coordinate axes are rotated through an angle of θ = 30°
Solution
We have x = 1, y = 2,
cos θ = cos 30° = ^{√
3}⁄₂,
sin θ = sin 30° = ¹⁄₂
Then
x' = x cos θ + y sin θ
x' = 1(^{√
3}⁄₂) + 2(¹⁄₂)
= ^{√
3}⁄₂ + 1
y' = - x sin θ + y cos θ
y' = -1(¹⁄₂) + 2(^{√
3}⁄₂)
= - ¹⁄₂ + √ 3.
Therefore, the new coordinates are (^{√
3}⁄₂ + 1, - ¹⁄₂ + √ 3)

eliminating the x'y'-Term

We have the general equation
Ax² + Bxy + Cy² + Dx + Ey + F = 0
with B ≠ 0.
We substitute
x = x' cos θ - y' sin θ and
y = x' sin θ + y' cos θ
in the general equation, then
A'(x')² + B'x'y' + C'(y')² + D'x' + E'y' + F' = 0
where
A' = A cos² θ + B cos θ sin θ + C sin² θ,
B' = B(cos² θ - sin² θ) + 2(C - A) sin θ cos θ,
= B(cos 2θ) - (A - C) sin 2θ,
C' = A sin θ - B sin θ cos θ + C cos² θ,
D = D cos θ + E sin θ,
E = -D sin θ + E cos θ,
F' = F.

To eliminate the x'y'-term, choose θ such that B' = 0.
So
B(cos 2θ) - (A - C) sin 2θ = 0
cot 2θ = ^{A - C}⁄_B.

Note
1. If cot 2θ = 0, then 2θ = 90° so that θ = 45° .
2. If cot 2θ ≠ 0, first find cos 2θ to obtain the correct acute angle θ.
Then we have

if B' = 0 , then cot 2θ = ^{A - C}⁄_B .

We can prove that
B'² - 4A'C' = B² - 4AC,
and since B' = 0, we have
B² - 4AC = - 4A'C' .
If B² - 4AC = 0, then -4 A'C' = 0 and we have
A' = 0 or C' = 0.
When A' = 0, C'y'² + D'x' + E'y' + F' = 0.
When C' = 0, A'x'² + D'x' + E'y' + F' = 0.
This means that if B² - 4AC = 0, after rotation with θ, then the general equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 becomes the standard form of the equation of a parabola.

The rotation of the coordinate axes through an angle θ transforms the equation
Ax² + Bxy + Cy² + Dx + Ey + F = 0
into the form
A'(x')² + B'x'y' + C'(y')² + D'x' + E'y' + F' = 0
has the following rotation invariants.
1. F = F'
2. A + C = A' + C'
3. B² - 4AC = B'² - 4A'C'

Example 13.
Rotate the coordinate axes to remove the xy-term of the equation
3x² - 2√ 3 xy + y² + 2x + 2√ 3 y = 0

in x'y'-coordinate system. Then sketch the graph.
Solution
3x² - 2√ 3 xy + y² + 2x + 2√ 3 y = 0

Comparing with the general equation, we get
A = 3, B = -2√ 3 , C = 1.
Therefore B² - 4AC = (-2√ 3 )² - 4(3)(1) = 0 and
cot 2θ = ^{A - C}⁄_B = ^{3 - 1}⁄_{-2√
3} = - ¹⁄_{√
3} .
Thus,
2θ = 120° and so θ = 60° .
Next,
sin θ = sin 60° = ^{√
3}⁄₂ and
cos θ = cos 60° = ¹⁄₂ .
By using x = x' cos θ - y' sin θ and
y = x' sin θ + y' cos θ, we get
x = ¹⁄₂x' - ^{√
3}⁄₂y' and
y = ^{√
3}⁄₂x' + ¹⁄₂y' .
By substituting x = ¹⁄₂x' - ^{√
3}⁄₂y' and
y = ^{√
3}⁄₂x' + ¹⁄₂y' in given equation, we get
³⁄₄((x')² - 2√ 3 x'y' + 3(y')²) - ^{√
3}⁄₂ (x' - √ 3 y')(√ 3 x' + y') + ¹⁄₄(3(x')² + 2√ 3 x'y' + (y')²) + x' - √ 3 y' + 3x' + √ 3 y' = 0
and then
(y')² = -x'.
We see that the new conic is a parabola with vertex (0, 0).

Example 14.
Rotate the coordinate axes to remove the xy-term of the equation
16x² - 24xy + 9y² + 100x - 200y + 100 = 0
in x'y'-coordinate system. Then sketch the graph.
Solution
16x² - 24xy + 9y² + 100x - 200y + 100 = 0
Comparing with the general equation, we get
A = 16, B = -24 , C = 9.
We find that B² - 4AC = (-24)² - 4(16)(9) = 0 and
cot 2θ = ^{A - C}⁄_B = ^{16 - 9}⁄_-24 = - ⁷⁄₂₄.
Using trigonometric identities, we see that
cos 2θ = - ⁷⁄₂₅.
Therefore,
sin θ = ⁴⁄₅ and cos θ = ³⁄₃.
We get x = ³⁄₅ x' - ⁴⁄₅y' and
y = ⁴⁄₅ x' + ³⁄₅y' .
Substituting these values into the given equation, we get
25(y')² - 100x' - 200y' + 100 = 0
(y')² - 8y' = 4x' - 4
(y' - 4)² = 4x' + 12
(y' - 4)² = 4(x' + 3).
We see that the new conic in x'y'-plane is a parabola with vertex (-3, 4).
Since sin θ = ⁴⁄₅ , we find that θ = 53.1° .

Exercise 6.4

Rotate the coordinate axes to remove the xy-term. Then sketch the graph.
1. x² + 2√ 3 xy + 3y² + 2√ 3 x - 2y = 0
Solution
x² + 2√ 3 xy + 3y² + 2√ 3 x - 2y = 0
Comparing with the general equation, we get
A = 1, B = 2√ 3 , C = 3
Therefore
B² = -4AC = (2√ 3 )² - 4(1)(3)
= 12 - 12 = 0
cot 2θ = ^{A - C}⁄_B = ^{1 - 3}⁄_{2√
3} = ^-2⁄_{√
3}
2θ = 120°
θ = 60°
sin θ = sin 60° = ^{√
3}⁄₂, cos θ = cos 60° = ¹⁄₂
By using x = x' cos θ - y' sin θ and
y = x' sin θ + y' cos θ,
x = ¹⁄₂ x' - ^{√
3}⁄₂ y' ,
y = ^{√ 3} ⁄₂ x' + ¹⁄₂ y'
By substituting x = ¹⁄₂ x' - ^{√
3}⁄₂ y' and
y = ^{√ 3} ⁄₂ x' + ¹⁄₂ y' in given equation,
(¹⁄₂ x' - ^{√
3}⁄₂ y')² + 2√ 3( ¹⁄₂ x' - ^{√ 3} ⁄₂ y') (^{√ 3} ⁄₂ x' + ¹⁄₂ y') + 3(^{√ 3} ⁄₂ x' + ¹⁄₂ y')² + 2√ 3( ¹⁄₂ x' - ^{√ 3} ⁄₂ y') - 2(^{√ 3} ⁄₂ x' + ¹⁄₂ y') = 0
¹⁄₄x'² - ^{√
3}⁄₂ x'y' + ³⁄₄ y'² + 2√ 3(^{√
3}⁄₄ x'² + ¹⁄₄ x'y' - ³⁄₄ x'y' - ^{√
3}⁄₄ y'²) + 3 (³⁄₄x'² + ^{√
3}⁄₂ x'y' + ¹⁄₄y'²) + √ 3 x' - 3 y' - √ 3 x' - y' = 0
¹⁄₄x'² - ^{√
3}⁄₂ x'y' + ³⁄₄ y'² + ³⁄₂x'² + ^{√ 3} ⁄₂ x'y' - ^{3√ 3} ⁄₂ x'y' - ³⁄₂ y'² + ⁹⁄₄x'² + ^{3√ 3} ⁄₂ x'y' + ³⁄₄y'² + √ 3 x' - 3y' - √ 3 x' - y' = 0
¹⁄₄x'² + ³⁄₄ y'² + ³⁄₂x'² - ³⁄₂ y'² + ⁹⁄₄x'² + ³⁄₄y'² - 4y' = 0
¹⁄₄x'² + ³⁄₄ y'² + ⁶⁄₄x'² - ⁶⁄₄ y'² + ⁹⁄₄x'² + ³⁄₄y'² - ¹⁶⁄₄y' = 0
¹⁄₄ (x'² + 3y'² + 6x'² - 6y'² + 9x'² + 3y'² - 16y') = 0
x'² + 3y'² + 6x'² - 6y'² + 9x'² + 3y'² - 16y' = 0
16x'² + 6y'² - 6y'² - 16y' = 0
16x'² - 16y' = 0
16(x'² - y') = 0
x'² - y' = 0
x'² = y'
(x' - 0)² = (y' + 0)
The new conic is a parabola with vertex (0, 0).

2. 9x² - 24xy + 16y² - 80x - 60y + 100 = 0
Solution
9x² - 24xy + 16y² - 80x - 60y + 100 = 0
Comparing with the general equation,
A = 9, B = -24 , C = 16
B² = -4AC = (-24)² - 4(9)(16)
= 576 - 576 = 0
cot 2θ = ^{A - C}⁄_B = ^{9 - 16}⁄_-24 = ^-7⁄_-24 = ⁷⁄₂₄

cos 2θ = ⁷⁄₂₅
cos 2θ = 2 cos² θ - 1
2 cos² θ - 1 = ⁷⁄₂₅
2 cos² θ = ⁷⁄₂₅ + 1
2 cos² θ = ³²⁄₂₅
cos² θ = ¹⁶⁄₂₅
cos θ = ⁴⁄₅
sin² θ + cos² θ = 1
sin² θ + ¹⁶⁄₂₅ = 1
sin² θ = 1 - ¹⁶⁄₂₅ = ⁹⁄₂₅
sin θ = ³⁄₅
x = x' cos θ - y' sin θ ,
y = x' sin θ + y' cos θ,
x = ⁴⁄₅x' - ³⁄₅y' ,
y = ³⁄₅x' + ⁴⁄₅y'
Substituting these values into the given equation,
9(⁴⁄₅x' - ³⁄₅y')² - 24 (⁴⁄₅x' - ³⁄₅y') (³⁄₅x' + ⁴⁄₅y') + 16 (³⁄₅x' + ⁴⁄₅y')² - 80 (⁴⁄₅x' - ³⁄₅y') - 60 (³⁄₅x' + ⁴⁄₅y') + 100 = 0
9(¹⁶⁄₂₅x'² - ²⁴⁄₂₅x'y' + ⁹⁄₂₅y'²) - 24 (¹²⁄₂₅x'² + ¹⁶⁄₂₅x'y' - ⁹⁄₂₅x'y' - ¹²⁄₂₅y'²) + 16 (⁹⁄₂₅x'² + ²⁴⁄₂₅x'y' + ¹⁶⁄₂₅y'²) - ³²⁰⁄₅x' + ²⁴⁰⁄₅y' - ¹⁸⁰⁄₅x' - ²⁴⁰⁄₅y' + 100 = 0
¹⁴⁴⁄₂₅x'² - ²¹⁶⁄₂₅x'y' + ⁸¹⁄₂₅y'² - ²⁸⁸⁄₂₅x'² - ³⁸⁴⁄₂₅x'y' + ²¹⁶⁄₂₅x'y' + ²⁸⁸⁄₂₅y'² + ¹⁴⁴⁄₂₅x'² + ³⁸⁴⁄₂₅x'y' + ²⁵⁶⁄₂₅y'² - ³²⁰⁄₅x' - ¹⁸⁰⁄₅x' + 100 = 0
¹⁴⁴⁄₂₅x'² + ⁸¹⁄₂₅y'² - ²⁸⁸⁄₂₅x'² + ²⁸⁸⁄₂₅y'² + ¹⁴⁴⁄₂₅x'² + ²⁵⁶⁄₂₅y'² - ³²⁰⁄₅x' - ¹⁸⁰⁄₅x' + 100 = 0
²⁸⁸⁄₂₅x'² + ⁸¹⁄₂₅y'² - ²⁸⁸⁄₂₅x'² + ²⁸⁸⁄₂₅y'² + ²⁵⁶⁄₂₅y'² - ³²⁰⁄₅x' - ¹⁸⁰⁄₅x' + 100 = 0
⁸¹⁄₂₅y'² + ²⁸⁸⁄₂₅y'² + ²⁵⁶⁄₂₅y'² - ³²⁰⁄₅x' - ¹⁸⁰⁄₅x' + 100 = 0
⁶²⁵⁄₂₅y'² - ⁵⁰⁰⁄₅x' + 100 = 0
25y'² - 100x' + 100 = 0
25(y'² -4x' + 4) = 0
y'² -4x' + 4 = 0
y'² = 4x' - 4
y'² = 4(x' - 1)
(y' - 0)² = 4(x' - 1)
The new conic in 'y'-plane is a parabola with vertex (1, 0).
Since sin θ = ³⁄₅, θ = 36.9°

3. x² - 2xy + y² - 20x - 20y + 100 = 0
Solution
x² - 2xy + y² - 20x - 20y + 100 = 0
Comparing with the general equation,
A = 1, B = -2 , C = 1
B² = -4AC = (-2)² - 4(1)(1)
= 4 - 4 = 0
cot 2θ = ^{A - C}⁄_B = ^{1 - 1}⁄_-2 = ⁰⁄_-2 = 0
2θ = 90°
θ = 45°
cos θ = cos 45° = ^{√
2}⁄₂ , sin θ = sin 45° = ^{√
2}⁄₂
x = x' cos θ - y' sin θ ,
y = x' sin θ + y' cos θ
x = ^{√
2}⁄₂ x' - ^{√
2}⁄₂ y'
y = ^{√
2}⁄₂ x' + ^{√
2}⁄₂ y'
Substituting these values into the given equation,
(^{√
2}⁄₂ x' - ^{√
2}⁄₂ y')² - 2(^{√
2}⁄₂ x' - ^{√
2}⁄₂ y') (^{√
2}⁄₂ y' + ^{√
2}⁄₂ x') + (^{√
2}⁄₂ y' + ^{√
2}⁄₂ x')² - 20(^{√
2}⁄₂ x' - ^{√
2}⁄₂ y') - 20(^{√
2}⁄₂ y' + ^{√
2}⁄₂ x') + 100 = 0
¹⁄₂x'² - x'y' + ¹⁄₂y'² - 2(¹⁄₂x'y' + ¹⁄₂x'² - ¹⁄₂y'² - ¹⁄₂x'y') + ¹⁄₂y'² + x'y' + ¹⁄₂x'² - 10√ 2 x' + 10√ 2 y'- 10√ 2 y' - 10√ 2 x' + 100 = 0
¹⁄₂x'² + ¹⁄₂y'² - 2(¹⁄₂x'y' + ¹⁄₂x'² - ¹⁄₂y'² - ¹⁄₂x'y') + ¹⁄₂y'² + ¹⁄₂x'² - 10√ 2 x' - 10√ 2 x' + 100 = 0
¹⁄₂ x'² + ¹⁄₂y'² - x'y' - x'² + y'² + x'y' + ¹⁄₂y'² + ¹⁄₂x'² - 20√ 2 x' + 100 = 0
x'² + y'² - x'² + y'² - 20√ 2 x' + 100 = 0
2y'² - 20 √ 2 x' + 100 = 0
2(y'² - 10√ 2 x' + 50) = 0
y'² - 10√ 2 x' + 50 = 0
y'² = 10√ 2 x' - 50
y'² = 10√ 2 x' - 5 ⋅ 10 ^{√
2}⁄_{√
2}
= 10√ 2 (x' - ⁵⁄_{√
2})
= 10√ 2 (x' - ^{5√
2}⁄₂)
The new conic in x'y'-plane is a parabola with (^{5 √
2}⁄₂ , 0)

4. 16x² + 24xy + 9y² - 130x + 90y = 0
Solution
16x² + 24xy + 9y² - 130x + 90y = 0
Comparing with the general equation,
A = 16 , B = 24 , C = 9
B² - 4AC = 24² - 4(16)(9)
= 576 - 576 = 0
cot 2θ = ^{A - C}⁄_B = ^{16 - 9}⁄₂₄ = ⁷⁄₂₄

cos 2θ = ⁷⁄₂₅
cos 2θ = 2 cos² θ - 1
2 cos² θ - 1 = ⁷⁄₂₅
2 cos² θ = ⁷⁄₂₅ + 1
2 cos² θ = ³²⁄₂₅
cos² θ = ¹⁶⁄₂₅
cos θ = ⁴⁄₅
sin² θ + cos² θ = 1
sin² θ + ¹⁶⁄₂₅ = 1
sin² θ = 1 - ¹⁶⁄₂₅ = ⁹⁄₂₅
sin θ = ³⁄₅
x = x' cos θ - y' sin θ ,
y = x' sin θ + y' cos θ,
x = ⁴⁄₅x' - ³⁄₅y' ,
y = ³⁄₅x' + ⁴⁄₅y'
Substituting these values into the given equation,
16(⁴⁄₅x' - ³⁄₅y')² + 24 (⁴⁄₅x' ³⁄₅y') (³⁄₅x' + ⁴⁄₅y') + 9 (³⁄₅x' + ⁴⁄₅y')² - 130 (⁴⁄₅x' - ³⁄₅y') + 90 (³⁄₅x' + ⁴⁄₅y') = 0
16(¹⁶⁄₂₅x'² - ²⁴⁄₂₅x'y' + ⁹⁄₂₅y'²) + 24 (¹²⁄₂₅x'² + ¹⁶⁄₂₅x'y' - ⁹⁄₂₅x'y' - ¹²⁄₂₅y'²) + 9 (⁹⁄₂₅x'² + ²⁴⁄₂₅x'y' + ¹⁶⁄₂₅y'²) - ⁵²⁰⁄₅x' + ³⁹⁰⁄₅y' + ²⁷⁰⁄₅x' + ³⁶⁰⁄₅y' = 0
²⁵⁶⁄₂₅x'² - ³⁸⁴⁄₂₅x'y' + ¹⁴⁴⁄₂₅y'² + ²⁸⁸⁄₂₅x'² + ³⁸⁴⁄₂₅x'y' - ²¹⁶⁄₂₅x'y' - ²⁸⁸⁄₂₅y'² + ⁸¹⁄₂₅x'² + ²¹⁶⁄₂₅x'y' + ¹⁴⁴⁄₂₅y'² - ²⁵⁰⁄₅x' + ⁷⁵⁰⁄₅y' = 0
²⁵⁶⁄₂₅x'² + ²⁸⁸⁄₂₅y'² + ²⁸⁸⁄₂₅x'² - ²⁸⁸⁄₂₅y'² + ⁸¹⁄₂₅x'² - 50x' + 150y' = 0
⁶²⁵⁄₂₅x'² - 50x' + 150y' = 0
25x'² - 25 ⋅ 2x' + 25 ⋅ 6y' = 0
25(x'² -2x' + 6y') = 0
x'² -2x' + 6y' = 0
x'² - 2x' = - 6y'
x'² - 2x' + 1 = -6y' + 1
(x' - 1)² = -6(y' - ¹⁄₆)
The new conic in x'y'-plane is a parabola with vertex (1, ¹⁄₆)

5. 3x² + 2√ 3 xy + y² - 8x + 8√ 3 y = 0

Solution
3x² + 2√ 3 xy + y² - 8x + 8√ 3 y = 0

Comparing with the general equation, we get
A = 3, B = 2√ 3 , C = 1
Therefore
B² = -4AC = (2√ 3 )² - 4(3)(1)
= 12 - 12 = 0
cot 2θ = ^{A - C}⁄_B = ^{3 - 1}⁄_{2√
3} = ¹⁄_{√
3}
2θ = 60°
θ = 30°
cos θ = cos 30° = ^{√
3}⁄₂, sin θ = sin 30° = ¹⁄₂
By using x = x' cos θ - y' sin θ and
y = x' sin θ + y' cos θ,
x = ^{√
3}⁄₂ x' - ¹⁄₂ y' ,
y = ¹⁄₂x' + ^{√ 3} ⁄₂ y'
By substituting x = ^{√
3}⁄₂ x' - ¹⁄₂y' and
y = ¹⁄₂x' + ^{√ 3} ⁄₂ y' in given equation,
3 (^{√
3}⁄₂ x' - ¹⁄₂y' )² + 2√ 3 (^{√ 3} ⁄₂ x' - ¹⁄₂y' ) (¹⁄₂x' + ^{√ 3} ⁄₂ y') + (¹⁄₂x' + ^{√ 3} ⁄₂ y')² -8(^{√ 3} ⁄₂ x' - ¹⁄₂y') + 8√ 3 (¹⁄₂ x' + ^{√ 3} ⁄₂ y') = 0
3(³⁄₄x'² - ^{√
3}⁄₂ x'y' + ¹⁄₄ y'²) + 2√ 3(^{√
3}⁄₄ x'² + ³⁄₄ x'y' - ¹⁄₄ x'y' - ^{√
3}⁄₄ y'²) + ¹⁄₄x'² + ^{√
3}⁄₂ x'y' + ³⁄₄y'² - 4√ 3 x' + 4y' + 4√ 3 x' + 12y' = 0
⁹⁄₄x'² - ^{3√
3}⁄₂ x'y' + ³⁄₄y'² + ³⁄₂x'² + ^{3√
3}⁄₂ x'y' - ^{√
3}⁄₂ x'y' - ³⁄₂ y'² + ¹⁄₄x'² + ^{√ 3} ⁄₂ x'y' + ³ ⁄₄ y'² + 16y' = 0
⁹⁄₄x'² + ³ ⁄₄ y'² + ⁶⁄₄x'² - ⁶⁄₄y'² + ¹⁄₄ x'² + ³⁄₄y'² + 16y' = 0
¹⁶⁄₄x'² + ⁶ ⁄₄ y'² - ⁶ ⁄₄ y'² + 16y' = 0
4x'² + 4⋅4y' = 0
4(x'² + 4y') = 0
x'² + 4y' = 0
x'² = -4y'
(x - 0)² = -4(y- 0)
The new conic is a parabola with vertex (0, 0).

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