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Topos Uranos
English

Rotation of axes and the general second-degree equation

Summary

This lesson completes the study of conics with the case that was left pending: the second-degree equation that contains a term in xyxy. Using polar coordinates and the difference formulas, it derives how the coordinates of a point change when the axes rotate through an angle θ\theta, and proves that the rotation preserves distances, so that it does not alter the shape of any figure. It computes the coefficients of the general equation after the rotation, and proves that the angle satisfying cot⁡2θ=A−CB\cot 2\theta = \frac{A - C}{B} eliminates the term in xyxy and that the sum A+CA + C and the discriminant B2−4ACB^2 - 4AC do not change under the rotation. From this follows the classification of all second-degree equations by the sign of the discriminant: ellipses, parabolas and hyperbolas, with their degenerate cases. It ends by recognizing as rotated conics the hyperbola xy=1xy = 1, an ellipse and a parabola whose axes are oblique.

Learning objectives

  1. Derive the formulas that relate the coordinates of a point in two Cartesian systems that differ by a rotation, and prove that the rotation preserves distances.
  2. Compute the coefficients of a second-degree equation after a rotation and choose the angle that eliminates the term in xyxy.
  3. Prove that A+CA + C and B2−4ACB^2 - 4AC are invariant under rotation and use them to obtain the reduced form without substituting.
  4. Classify a general second-degree equation by the sign of its discriminant, find its elements in the rotated coordinates and express them in the original coordinates.

The rotation of the axes

In the Course I lesson on conic sections it was proved that every second-degree equation without a term in xyxy represents a parabola, an ellipse, a hyperbola or one of their degenerate cases, whose axes of symmetry are parallel to the coordinate axes. Now, an ellipse whose major axis lies on the line y=xy = x also has a second-degree equation, and a term in xyxy appears in it. To recognize it, it is convenient to choose another system of axes, rotated with respect to the first, in which the curve recovers its canonical form. We now study how the coordinates of a point change when the axes are rotated.

Consider a Cartesian system with axes xx and yy, like those of the Course I lesson on the equation of a line and Cartesian coordinate systems, and let θ\theta be a real number. The system rotated by the angle θ\theta is the Cartesian system that has the same origin and the same unit of length, whose positive semi-axis of abscissas is the ray starting from the origin with angle θ\theta in standard position, and whose positive semi-axis of ordinates is the ray of angle θ+π2\theta + \frac{\pi}{2}. The new coordinates are written with capital letters: one and the same point PP has coordinates (x,y)(x, y) in the original system and (X,Y)(X, Y) in the rotated one. Since the two systems have the same orientation, polar coordinates are also defined in the rotated system, with the same pole and with the positive semi-axis of XX as the polar axis.

TheoremThe rotation formulas

If a point has coordinates (x,y)(x, y) in a Cartesian system and coordinates (X,Y)(X, Y) in the system rotated by the angle θ\theta, then its new coordinates are obtained from the original ones, and the original ones from the new ones, by means of the formulas below.

X=xcos⁡θ+ysin⁡θ,Y=−xsin⁡θ+ycos⁡θX = x \cos \theta + y \sin \theta, \qquad Y = -x \sin \theta + y \cos \theta
x=Xcos⁡θ−Ysin⁡θ,y=Xsin⁡θ+Ycos⁡θx = X \cos \theta - Y \sin \theta, \qquad y = X \sin \theta + Y \cos \theta

The idea is that, in polar coordinates, rotating the axes does not change the distance to the pole and only subtracts θ\theta from the polar angle; the sum and difference formulas, which are proved in the lesson on the sum and difference formulas, do the rest.

Proof

  1. x=rcos⁡φ,y=rsin⁡φ\dato{xy}{x = r \cos \varphi, \quad y = r \sin \varphi}

    If PP is the origin, its coordinates are zero in both systems and the formulas hold. If it is not, we take a polar representation (r,φ)(r, \varphi) with r>0r > 0 with respect to the original polar axis, as in the lesson on polar coordinates.

  2. X=rcos⁡(φ−θ),Y=rsin⁡(φ−θ)\dato{XY}{X = r \cos(\varphi - \theta), \quad Y = r \sin(\varphi - \theta)}

    In the rotated system the pole is the same, so the distance from PP to the pole is still rr. Let ψ\psi be a polar angle of PP in the rotated system. On the unit circle, the new polar axis passes through P(θ)P(\theta); traversing an arc ψ\psi from there leads, by the concatenation property of the axiom of traversal from the lesson on angles and their measure, to P(θ+ψ)P(\theta + \psi), which must be the point P(φ)P(\varphi) of the ray that goes from the pole to PP. By the coterminal angles lemma, θ+ψ\theta + \psi and φ\varphi differ by a multiple of 2π2\pi, so we can take ψ=φ−θ\psi = \varphi - \theta.

  3. X=rcos⁡(φ−θ)=rcos⁡φcos⁡θ+rsin⁡φsin⁡θ=xcos⁡θ+ysin⁡θX = \resaltar{r \cos(\varphi - \theta)} = r \cos \varphi \cos \theta + r \sin \varphi \sin \theta = \resaltar{x \cos \theta + y \sin \theta}

    We expand the cosine of the difference and recognize the original coordinates in each term.

  4. Y=rsin⁡(φ−θ)=rsin⁡φcos⁡θ−rcos⁡φsin⁡θ=−xsin⁡θ+ycos⁡θY = \resaltar{r \sin(\varphi - \theta)} = r \sin \varphi \cos \theta - r \cos \varphi \sin \theta = \resaltar{-x \sin \theta + y \cos \theta}

    Likewise, with the sine of the difference.

  5. x=rcos⁡((φ−θ)+θ)=Xcos⁡θ−Ysin⁡θ,y=rsin⁡((φ−θ)+θ)=Xsin⁡θ+Ycos⁡θx = \resaltar{r \cos((\varphi - \theta) + \theta)} = \resaltar{X \cos \theta - Y \sin \theta}, \qquad y = \resaltar{r \sin((\varphi - \theta) + \theta)} = \resaltar{X \sin \theta + Y \cos \theta}

    Conversely, the original angle is the sum φ=(φ−θ)+θ\varphi = (\varphi - \theta) + \theta; the sum formulas give the original coordinates in terms of the new ones.

The two pairs of formulas have the same structure: the second pair is obtained from the first by replacing θ\theta with −θ-\theta, because the original system is the one that results from rotating the new one by the angle −θ-\theta. For example, with θ=π4\theta = \frac{\pi}{4}, the point (1,1)(1, 1) has new coordinates X=2X = \sqrt{2} and Y=0Y = 0: it lies on the new axis of abscissas, at distance 2\sqrt{2} from the origin. One should not confuse the rotation of the axes with the rotation of the points: rotating the axes by the angle θ\theta leaves the points fixed and changes their names, and the result is the same as leaving the axes fixed and rotating the whole plane by the angle −θ-\theta.

LemmaRotation preserves distances

The distance between two points, computed with their coordinates in the rotated system, coincides with the one computed with their original coordinates.

The fact was already observed in the lesson on polar coordinates: the distance between two points depends on their polar angles only through their difference, and rotating the plane amounts to adding a constant to those angles. Here it is checked directly in Cartesian coordinates, with the rotation formulas and the Pythagorean identity from the lesson on trigonometric ratios.

Proof

  1. X1−X2=ucos⁡θ+vsin⁡θ,Y1−Y2=−usin⁡θ+vcos⁡θ\dato{dif}{X_1 - X_2 = u \cos \theta + v \sin \theta, \quad Y_1 - Y_2 = -u \sin \theta + v \cos \theta}

    Let P1P_1 and P2P_2 be two points, with coordinates (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) in the original system and (X1,Y1)(X_1, Y_1) and (X2,Y2)(X_2, Y_2) in the rotated one. We subtract the rotation formulas written for each point, with u=x1−x2u = x_1 - x_2 and v=y1−y2v = y_1 - y_2.

  2. (X1−X2)2+(Y1−Y2)2=(ucos⁡θ+vsin⁡θ)2+(−usin⁡θ+vcos⁡θ)2=u2(cos⁡2θ+sin⁡2θ)+v2(sin⁡2θ+cos⁡2θ)(X_1 - X_2)^2 + (Y_1 - Y_2)^2 = \resaltar{(u \cos \theta + v \sin \theta)^2 + (-u \sin \theta + v \cos \theta)^2} = \resaltar{u^2\left(\cos^2 \theta + \sin^2 \theta\right) + v^2\left(\sin^2 \theta + \cos^2 \theta\right)}

    We square and add. The cross terms, 2uvsin⁡θcos⁡θ2uv \sin \theta \cos \theta and −2uvsin⁡θcos⁡θ-2uv \sin \theta \cos \theta, cancel.

  3. (X1−X2)2+(Y1−Y2)2=u2+v2=(x1−x2)2+(y1−y2)2(X_1 - X_2)^2 + (Y_1 - Y_2)^2 = \resaltar{u^2 + v^2} = (x_1 - x_2)^2 + (y_1 - y_2)^2

    By the Pythagorean identity, each parenthesis equals 11, and what remains is the square of the original distance.

The lemma has a consequence that justifies everything that follows. The parabola, the ellipse and the hyperbola were defined in the earlier lessons by properties of distances to foci and directrices; since the rotation preserves distances, a curve that in the rotated system is an ellipse with semi-axes aa and bb is, seen in the original system, an ellipse with the same semi-axes, even though its axes of symmetry are oblique. The same holds for the other conics, for their foci, their vertices and their asymptotes, which are carried from one system to the other with the rotation formulas.