SME
Grade 12 Math · Problem type
The equation after turning the axes
Use this for “Turn the axes to remove the xy term of 4x² + 2xy + 2y² = 1 and write the equation in x′ and y′.”
Diagram
The
turn
has
tan
2θ
=
2
4 − 2
=
1,
so
2θ
=
45°
and
θ
=
22.5°.
In the turned axes the equation reads 4.41x′² + 1.59y′² = 1, with no x′y′ term: an ellipse, as B² − 4AC = −28 < 0 says.
Its
half-axes
are
√(
1
A′
)
=
0.48
along
x′
and
√(
1
C′
)
=
0.79
along
y′.
This is an example. Type your own values to start a new problem.
A
x²
+
B
xy
+
C
y²
+
(
F
)
=
0
A′
x′²
+
C′
y′²
+
(
F
)
=
0
θ
Angle to turn the axes
worked out (not typed) · °
Formulas
tan(2
×
θ
)
=
B
A
−
C
tan(2
×
22.5°)
=
2
4 − 2
A
′
=
A
×
cos
(
θ
)
2
+
B
×
sin(
θ
)
×
cos(
θ
)
+
C
×
sin
(
θ
)
2
4.4142
=
4
×
cos
(22.5°)
2
+
2
×
sin(22.5°)
×
cos(22.5°)
+
2
×
sin
(22.5°)
2
C
′
=
A
×
sin
(
θ
)
2
−
B
×
sin(
θ
)
×
cos(
θ
)
+
C
×
cos
(
θ
)
2
1.5858
=
4
×
sin
(22.5°)
2
−
2
×
sin(22.5°)
×
cos(22.5°)
+
2
×
cos
(22.5°)
2
Assumptions
•
Turning by θ puts x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ into the equation.
•
With tan 2θ = B ÷ (A − C) the x′y′ term is 0, leaving A′x′² + C′y′² = −F.
•
A′ + C′ = A + C: a quick check on the arithmetic.
•
At θ = 22.5°, 30° or 45° the numbers have exact √ forms (3 + √2); the page shows 4 decimal places.
Step-by-step
WE KNOW
A = 4, B = 2, C = 2, F = −1
FIND
angle to turn the axes (θ), number before x′² (A′), number before y′² (C′)
Step 1 · Find angle to turn the axes (θ)
Use tan(2 × θ) = B ÷ (A − C)
tan
2
θ
=
B
A
−
C
is
cot
2
θ
=
A
−
C
B
turned
over;
add
180°
to
a
negative
2
θ
,
then
halve.
θ
=
tan⁻¹
B
÷ (
A
−
C
)
2
θ
=
tan⁻¹
2 ÷ (4 − 2)
2
θ
=
45°
÷
2
θ
=
22.5°
Step 2 · Find number before x′² (A′)
Use A′ = A × cos(θ)² + B × sin(θ) × cos(θ) + C × sin(θ)²
Put
x
=
x′
cos
θ
−
y′
sin
θ
and
y
=
x′
sin
θ
+
y′
cos
θ
in,
and
collect
the
x′²
terms.
A
′
=
4
×
cos
(22.5°)
2
+
2
×
sin(22.5°)
×
cos(22.5°)
+
2
×
sin
(22.5°)
2
A
′
=
4
×
0.8536
+
2
×
0.3536
+
2
×
0.1464
A
′
=
3.4142
+
0.7071
+
0.2929
A
′
=
4.4142
Step 3 · Find number before y′² (C′)
Use C′ = A × sin(θ)² − B × sin(θ) × cos(θ) + C × cos(θ)²
Collect the y′² terms the same way; the signs of the sines change.
C
′
=
4
×
sin
(22.5°)
2
−
2
×
sin(22.5°)
×
cos(22.5°)
+
2
×
cos
(22.5°)
2
C
′
=
4
×
0.1464
−
2
×
0.3536
+
2
×
0.8536
C
′
=
0.5858
−
0.7071
+
1.7071
C
′
=
1.5858
→
4.4142x′²
+
1.5858y′²
=
1,
with
A
′
+
C
′
=
A
+
C
Check
Put the numbers back in. Both sides should match.
tan(2
×
22.5°)
=
2
4 − 2
✓
4.4142
≈
4
×
cos
(22.5°)
2
+
2
×
sin(22.5°)
×
cos(22.5°)
+
2
×
sin
(22.5°)
2
✓
1.5858
≈
4
×
sin
(22.5°)
2
−
2
×
sin(22.5°)
×
cos(22.5°)
+
2
×
cos
(22.5°)
2
✓
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