WorkbenchSlowly drag the angle θ and watch the reading B′; find the moment it becomes 0. Meanwhile, check whether the two readings below it ever move.
Loading figure…
What this move does
The origin stays put and both coordinate axes turn together about the origin through an angle θ. The new and old coordinates of a point satisfy
x=x′cosθ−y′sinθ,y=x′sinθ+y′cosθ.
Substitute into Ax2+Bxy+Cy2+Dx+Ey+F=0, and the new coefficient of x′y′ is
B′=Bcos2θ−(A−C)sin2θ.
Choose the angle: set B′=0, that is, cot2θ=BA−C, and take 0<θ<2π. If A=C, take θ=45∘ straight away.
Substitute: compute A′,C′,D′,E′ (F does not change).
Then translate: hand any remaining linear terms over to the previous stop.
Predict first
To remove the xy-term from xy=2, through how many degrees should the axes turn?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
A=C=0, so cot2θ=0, 2θ=90∘, θ=45∘.
Substitute x=2x′−y′, y=2x′+y′: xy=2x′2−y′2=2, that is, x′2−y′2=4. So the graph of an inverse-proportion function is a rectangular hyperbola turned through 45∘.
Loading figure…
xy=2: turn to 45∘ and see.
Work through one
Simplify 5x² + 4xy + 2y² − 6 = 0
Step 1 / 5
Loading figure…
The original equation
A=5, B=4, C=2. B2−4AC=16−40=−24<0: elliptic type. The figure shows a tilted ellipse.
5x2+4xy+2y2−6=0
Turn by some angle
Compute the right angle
Read off the new equation
The invariants
Why does A+C not change? Write the quadratic part as Ax2+Bxy+Cy2. Its value depends only on the point, not on the coordinate system. Evaluate it at two perpendicular unit directions and add: you get exactly A+C (in the new system, A′+C′), and this does not depend on which pair of perpendicular directions you take. The invariance of B2−4AC can be checked by substituting the formulas directly.
Practice
Exercise 1Basic
To remove the xy-term of x2+3xy+2y2−4=0, through what angle θ (0∘<θ<90∘) should the axes turn? Give it in degrees.
Hint · direction
Use cot2θ=BA−C.
Hint · method
A=1, B=3, C=2.
Hint · first step
cot2θ=−31, so 2θ is in the second quadrant.
Full solution
2θ=120∘, θ=60∘.
Exercise 2Basic
Rotating the axes turns 3x2+2xy+3y2=8 into A′x′2+C′y′2=8. What is the larger of A′,C′?
Hint · direction
You do not have to actually rotate: use the invariants.
Hint · method
A′+C′=A+C and −4A′C′=B2−4AC.
Hint · first step
A′+C′=6, A′C′=436−4=8.
Full solution
A′,C′ are the two roots 4 and 2 of t2−6t+8=0; the larger is 4. (Turning by 45∘ gives 4x′2+2y′2=8.)
Exercise 3Advanced
The curve 2x2−4xy+5y2=6 is an ellipse. Find the square of its semi-major axis.
Hint · direction
After rotating it is A′x′2+C′y′2=6; find A′,C′ with the invariants.
Hint · method
A′+C′=7, A′C′=44⋅2⋅5−16=6.
Hint · first step
A′,C′ are 1 and 6, so the equation is x′2+6y′2=6.
Full solution
6x′2+y′2=1, so the square of the semi-major axis is 6.
Exercise 4Challenge
The focus of the hyperbola xy=8 in the first quadrant is (m,n). Find m.
Hint · direction
First turn by 45∘ to get the standard form.
Hint · method
xy=2x′2−y′2, which gives x′2−y′2=16.
Hint · first step
a2=b2=16, c2=32, c=42; the foci lie on the x′-axis, that is, on the line y=x, at distance 42 from the origin.
The wrong step: To remove the xy-term of 5x2+4xy+2y2−6=0: cot2θ=BC−A=−43, θ≈63.4∘.
What went wrong? Think first, then open
DiagnosisSubstitute back into B′=Bcos2θ−(A−C)sin2θ: cos2θ=−0.6, sin2θ=0.8, so B′=4×(−0.6)−3×0.8=−4.8=0. With the sign flipped, the xy-term actually got bigger.
✓
The right step: B′=0 gives cot2θ=BA−C=43, θ≈26.6∘. If in doubt, check with the formula for B′.
“cot 2θ = (A − C)/B, with A first”
Simplifying by Rotating the Axes ← → step · F full screen · Esc exit