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📏Honors Pre-Calculus Unit 10 Review

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10.4 Rotation of Axes

10.4 Rotation of Axes

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
📏Honors Pre-Calculus
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Conic Sections and Rotation of Axes

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General form of conic sections

Every conic section can be written in the general second-degree equation:

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

The coefficients AA, BB, and CC tell you which type of conic you're dealing with. The key factor is whether the BxyBxy term is present, because that term is what indicates rotation.

When B=0B = 0 (no rotation):

  • A=CA = C: circle
  • A=0A = 0 or C=0C = 0 (but not both): parabola
  • AA and CC have the same sign but ACA \neq C: ellipse
  • AA and CC have opposite signs: hyperbola

When B0B \neq 0, the conic is rotated relative to the coordinate axes. You can still classify it using the discriminant B24ACB^2 - 4AC:

  • B24AC<0B^2 - 4AC < 0: ellipse (or circle)
  • B24AC=0B^2 - 4AC = 0: parabola
  • B24AC>0B^2 - 4AC > 0: hyperbola

The discriminant is invariant under rotation, meaning it gives the same value no matter how the axes are oriented. That's what makes it so useful for classification.

General form of conic sections, Rotation of Axes · Algebra and Trigonometry

Rotation of axes for conics

The whole point of rotating axes is to eliminate the BxyBxy term. Once that term is gone, you're left with a standard-looking conic equation that's much easier to analyze.

Finding the rotation angle:

The angle θ\theta that eliminates the xyxy term satisfies:

cot(2θ)=ACB\cot(2\theta) = \frac{A - C}{B}

This is equivalent to θ=12arctan ⁣(BAC)\theta = \frac{1}{2}\arctan\!\left(\frac{B}{A - C}\right) when ACA \neq C. If A=CA = C, then cot(2θ)=0\cot(2\theta) = 0, which gives θ=45°\theta = 45°.

Rotation formulas (original coordinates in terms of rotated coordinates):

x=xcosθysinθx = x'\cos\theta - y'\sin\theta

y=xsinθ+ycosθy = x'\sin\theta + y'\cos\theta

These express the old (x,y)(x, y) coordinates in terms of the new rotated (x,y)(x', y') system.

Step-by-step process for eliminating the xyxy term:

  1. Start with the general equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

  2. Compute θ\theta using cot(2θ)=ACB\cot(2\theta) = \frac{A - C}{B}

  3. Find cosθ\cos\theta and sinθ\sin\theta (use the half-angle identities if needed)

  4. Substitute the rotation formulas for xx and yy into the original equation

  5. Expand and simplify. The xyx'y' terms will cancel, leaving:

Ax2+Cy2+Dx+Ey+F=0A'x'^2 + C'y'^2 + D'x' + E'y' + F' = 0

The new coefficients AA', CC', DD', EE', and FF' will generally differ from the originals, but the BxyB'xy' term is now zero.

General form of conic sections, Conic Sections | That Which We Have Heard & Known

Standard form of rotated conics

After rotation eliminates the xyxy term, you still may need to complete the square to reach true standard form.

Completing the square on the rotated equation:

Starting from Ax2+Cy2+Dx+Ey+F=0A'x'^2 + C'y'^2 + D'x' + E'y' + F' = 0, group and complete the square:

A ⁣(x+D2A)2+C ⁣(y+E2C)2=F+D24A+E24CA'\!\left(x' + \frac{D'}{2A'}\right)^2 + C'\!\left(y' + \frac{E'}{2C'}\right)^2 = -F' + \frac{D'^2}{4A'} + \frac{E'^2}{4C'}

(This form applies when both AA' and CC' are nonzero. For a parabola, one of them is zero, so you only complete the square on the surviving squared variable.)

You can then translate to center the conic at the origin by setting x=x+D2Ax'' = x' + \frac{D'}{2A'} and y=y+E2Cy'' = y' + \frac{E'}{2C'}.

Standard forms by conic type:

  • Ellipse: x2a2+y2b2=1\frac{x''^2}{a^2} + \frac{y''^2}{b^2} = 1
  • Hyperbola: x2a2y2b2=1\frac{x''^2}{a^2} - \frac{y''^2}{b^2} = 1
  • Parabola: y=ax2y'' = ax''^2 or x=ay2x'' = ay''^2 (only one variable is squared)

Direct analysis of conic equations

Once you've rotated and translated to standard form, you can read off the key features directly.

Ellipse (x2a2+y2b2=1\frac{x''^2}{a^2} + \frac{y''^2}{b^2} = 1, with a>ba > b):

  • Center at the origin of the translated/rotated system
  • Vertices at (±a,0)(\pm a, 0) along the major axis and (0,±b)(0, \pm b) along the minor axis
  • Foci at (±c,0)(\pm c, 0), where c2=a2b2c^2 = a^2 - b^2
  • The defining property: the sum of distances from any point on the ellipse to the two foci is constant (equal to 2a2a)

Hyperbola (x2a2y2b2=1\frac{x''^2}{a^2} - \frac{y''^2}{b^2} = 1):

  • Center at the origin of the translated/rotated system
  • Vertices at (±a,0)(\pm a, 0) along the transverse axis
  • Foci at (±c,0)(\pm c, 0), where c2=a2+b2c^2 = a^2 + b^2
  • The defining property: the absolute difference of distances from any point on the hyperbola to the two foci is constant (equal to 2a2a)

Parabola (e.g., x=14py2x'' = \frac{1}{4p}y''^2):

  • Vertex at the origin of the translated/rotated system
  • Focus at distance pp from the vertex along the axis of symmetry
  • Directrix is a line perpendicular to the axis of symmetry, at distance pp on the opposite side of the vertex from the focus

To convert these features back to the original (x,y)(x, y) coordinate system, apply the rotation formulas in reverse and then undo the translation.

Advanced rotation concepts

  • Matrix form: The rotation can be written as a matrix equation. The rotation matrix (cosθsinθsinθcosθ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} maps (x,y)(x', y') coordinates to (x,y)(x, y) coordinates.
  • Quadratic forms: The second-degree terms Ax2+Bxy+Cy2Ax^2 + Bxy + Cy^2 can be expressed as xTMx\mathbf{x}^T M \mathbf{x}, where M=(AB/2B/2C)M = \begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}. The eigenvalues of MM give the coefficients AA' and CC' in the rotated equation, and the eigenvectors point along the principal axes of the conic.
  • Principal axes: These are the directions along which the conic has its maximum and minimum curvature. Finding them through eigenvalue analysis is equivalent to finding the rotation angle θ\theta algebraically.
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