Mathematics.

conic sections

Rotation of Conics

Analytic Geometry30 minDifficulty6 out of 10

You should know: conic sections

Overview

When the general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 has a nonzero xy-term (B ≠ 0), the conic it describes is tilted relative to the coordinate axes. Rotating the coordinate system by a suitable angle θ eliminates the xy-term, transforming the equation into the familiar axis-aligned standard form, from which the conic's type and orientation become immediately visible. The rotation angle is determined by cot(2θ) = (A−C)/B, and the discriminant B²−4AC is invariant under rotation, so it classifies the conic's type both before and after rotating.

Intuition

The xy-term in the general conic equation exists precisely because the conic's axes of symmetry are tilted relative to the x- and y-axes — if you could align your coordinate grid with the conic's own natural axes, the cross term would vanish and you'd be left with a simple standard-form equation. Rotating the coordinate system by the correct angle θ does exactly that: it's not the conic that moves, but the grid we measure it against, until the grid lines match the conic's own symmetry directions. Because rotation doesn't stretch or reflect the curve, the fundamental shape classification (ellipse/parabola/hyperbola) can't change, which is exactly why the discriminant B²−4AC stays the same before and after.

Formal Definition

Definition

To eliminate the xy-term from Ax²+Bxy+Cy²+Dx+Ey+F=0, rotate the axes by angle θ using:

x=x′cos⁡θ−y′sin⁡θ,y=x′sin⁡θ+y′cos⁡θx = x'\cos\theta - y'\sin\theta, \qquad y = x'\sin\theta + y'\cos\theta
Rotation of coordinates
cot⁡(2θ)=A−CB(B≠0)\cot(2\theta) = \frac{A-C}{B} \quad (B \neq 0)
Angle that eliminates the xy-term
B2−4AC=B′2−4A′C′B^2 - 4AC = B'^2 - 4A'C'
Discriminant is invariant under rotation

Worked Examples

  1. Here A=0, B=1, C=0, so the discriminant is positive, indicating a hyperbola.

    Δ=B2−4AC=1−0=1>0  ⟹  hyperbola\Delta = B^2-4AC = 1-0 = 1 > 0 \implies \text{hyperbola}

  2. Find the rotation angle: cot(2θ) = (A−C)/B = 0, so 2θ=90°, θ=45°.

    θ=45°\theta = 45°

  3. Substitute x = (x'−y')/√2, y = (x'+y')/√2 into xy=1 and simplify.

    xy=(x′)2−(y′)22=1  ⟹  x′22−y′22=1xy = \frac{(x')^2-(y')^2}{2} = 1 \implies \frac{x'^2}{2} - \frac{y'^2}{2} = 1

Answer: xy=1 becomes x'²/2 − y'²/2 = 1 in the rotated frame — a hyperbola with vertices at (x',y')=(±√2, 0).

Practice Problems

Difficulty 6/10

Compute the discriminant of 3x² − 2xy + 3y² − 8 = 0 and classify the conic.

Difficulty 7/10

For 3x² − 2xy + 3y² − 8 = 0, find the rotation angle θ that eliminates the xy-term.

Difficulty 8/10

An elliptical satellite dish's cross-section is tilted 30° from a design template modeled by x²/9+y²/4=1. Explain (qualitatively, using the rotation formulas) why the manufactured dish's equation in the original (unrotated) coordinate frame will contain a nonzero xy-term.

Quiz

The angle θ that eliminates the xy-term in Ax²+Bxy+Cy²+Dx+Ey+F=0 satisfies:
Under a rotation of axes, the discriminant B²−4AC:
Rotating xy=1 by 45° transforms it into:

Summary

  • A nonzero xy-term in the general conic equation means the conic is tilted relative to the coordinate axes.
  • Rotating by θ where cot(2θ)=(A−C)/B eliminates the xy-term and reveals the conic's standard form.
  • The discriminant B²−4AC is invariant under rotation, so it classifies the conic's type in any coordinate frame.

References

  1. WebsiteWikipedia — Conic section (rotation of axes)