Rotated Conics and Their Equations

The four conics, line, parabola, ellipse, hyperbola, have nice equations only when their axes line up with the coordinate axes. What if a conic is tilted? In its general form $A x^2 + Bxy + C y^2 + Dx + Ey + F = 0$, the offender is the cross term $Bxy$. When $B \neq 0$, the conic's axes are rotated relative to the $x$- and $y$-axes. The fix: rotate the coordinate system by the right angle so the cross term vanishes, then identify the conic.

The discriminant decides the type

The quantity $\Delta = B^2 - 4AC$ is invariant under rotation. It classifies the conic before any rotating: $\Delta < 0$ ellipse (or circle); $\Delta = 0$ parabola; $\Delta > 0$ hyperbola.

The rotation that kills the cross term

Let new coordinates $(x', y')$ relate to old by rotation through angle $\theta$: $x = x' \cos\theta - y' \sin\theta$, $y = x' \sin\theta + y' \cos\theta$. The new cross-term coefficient is $B' = B \cos 2\theta + (C - A) \sin 2\theta$. Set $B' = 0$: $\tan 2\theta = B/(A - C)$ when $A \neq C$, else $\theta = \pi/4$.

• Computer Graphics: Detecting whether a tilted ellipse fits a set of points (e.g., fitting a galaxy's shape to a star catalogue) needs the $xy$ cross term; rotating the axes to eliminate it gives the principal axes, basis of principal component analysis.
• Statistics: Principal Component Analysis (PCA) IS the rotation that eliminates the $xy$ term in a 2D covariance ellipse; the new axes are the directions of maximum variance, which is why PCA is the most-taught dimensionality-reduction method.
• Materials Engineering: Stress tensors in materials science have off-diagonal $xy$ terms in arbitrary coordinates; engineers rotate to principal-stress axes (the same rotation you are computing here) to find where a beam will crack first.

(Select "hyperbola" mode and slide the rotation control. Watch the asymptotes tilt with the conic.)

Worked example: xy = 1

For $xy = 1$, $A = C = 0$, so $\theta = \pi/4$. Substitute $x = (x' - y')/\sqrt{2}$, $y = (x' + y')/\sqrt{2}$: $xy = ((x')^2 - (y')^2)/2 = 1$, giving $(x')^2/2 - (y')^2/2 = 1$, a standard hyperbola.

Try it

• Classify $2x^2 + 3xy - 2y^2 = 5$ using the discriminant.
• For $3x^2 + 2xy + 3y^2 = 8$, find the rotation angle and the type.
• Classify $x^2 - 2xy + y^2 = 4$.

Pause: why is $B^2 - 4AC$ invariant under rotation? Because rotation is a rigid motion of the plane and the type of conic is a geometric property that cannot change under rigid motion.

A trap to watch for

The angle in $\tan 2\theta = B/(A - C)$ is $2\theta$, not $\theta$. Many beginners write $\tan\theta = B/(A - C)$ and get an angle twice the correct one. Always remember the factor of $2$. A second trap: after rotating, the cross term vanishes only if you used the right $\theta$, verify by substituting back.

What you now know

You can recognise a general second-degree equation, compute the discriminant, find the rotation angle that eliminates the cross term, apply the rotation, and identify the conic. This closes Chapter 12; the next chapter begins the study of functions.

Quick check

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References

• Lang, S. (1971). Basic Mathematics. Springer. Chapter 12, §5, rotating axes to standardise general second-degree equations.
• Apostol, T. M. (1969). Calculus, Volume 2. Wiley. Chapter 13: invariants of conics under rotation.
• Hartshorne, R. (2000). Geometry: Euclid and Beyond. Springer. Chapter 7: projective treatment of conics.