Unit 8, lesson 96

Rotation of Axes

How turning the coordinate axes changes the coordinates of points, why a tilted conic has an xy term, how to choose the angle that removes it, and what the discriminant can and cannot tell you.

  • Theory part 191
  • Practice part 192
  • Conic Sections

The equation xy=2xy = 2 describes a hyperbola, yet it matches none of the standard forms, because the curve is tilted. Its axes of symmetry are the diagonal lines y=xy = x and y=−xy = -x. Instead of turning the curve, you can turn the coordinate grid underneath it until the grid lines up with the curve. In the turned coordinates the tilted conic gets a familiar standard equation again.

Key terms

Geometric rotation

Definition The rotation of the plane about the origin through the angle θ\theta, counterclockwise when θ>0\theta > 0, sends each point (x,y)(x, y) to

(xcos⁡θ−ysin⁡θ,  xsin⁡θ+ycos⁡θ).(x\cos\theta - y\sin\theta, \; x\sin\theta + y\cos\theta).

To see why, write the point as (rcos⁡φ,rsin⁡φ)(r\cos\varphi, r\sin\varphi). The rotation adds θ\theta to the angle, so the image is (rcos⁡(φ+θ),rsin⁡(φ+θ))(r\cos(\varphi + \theta), r\sin(\varphi + \theta)), and the sum formulas expand these into the expression above. A rotation keeps all distances and angles. If a curve is the solution set of G(x,y)=0G(x, y) = 0, the curve rotated through θ\theta is the solution set of

G(xcos⁡θ+ysin⁡θ,−xsin⁡θ+ycos⁡θ)=0,\begin{aligned} G(&x\cos\theta + y\sin\theta, \\[0.4em] &{-x\sin\theta} + y\cos\theta) = 0, \end{aligned}

because a point lies on the rotated curve exactly when rotating it back through −θ-\theta lands on the original curve.

In plain words Turn the whole picture around the origin like a wheel. The sum formulas for sine and cosine turn that motion into formulas for the coordinates.

Example Rotate P=(3,1)P = (\sqrt{3}, 1) through θ=π6\theta = \frac{\pi}{6}. The point is 22 units from the origin at the angle π6\frac{\pi}{6}, so its image is 22 units out at the angle π3\frac{\pi}{3}, which is the point (1,3)(1, \sqrt{3}). The formula agrees, since

3⋅32−1⋅12=1and3⋅12+1⋅32=3.\begin{gathered} \sqrt{3} \cdot \frac{\sqrt{3}}{2} - 1 \cdot \frac{1}{2} = 1 \\[0.8em] \text{and} \\[0.4em] \sqrt{3} \cdot \frac{1}{2} + 1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}. \end{gathered}

Rotated coordinates

Definition Turn the xx-axis and the yy-axis counterclockwise about the origin through the angle θ\theta, and call the turned axes the XX-axis and the YY-axis. Every point stays where it is but gets new coordinates (X,Y)(X, Y), measured along the new axes. Old and new coordinates are related by

x=Xcos⁡θ−Ysin⁡θ,y=Xsin⁡θ+Ycos⁡θ,\begin{aligned} x &= X\cos\theta - Y\sin\theta, \\[0.4em] y &= X\sin\theta + Y\cos\theta, \end{aligned}

and, solved the other way,

X=xcos⁡θ+ysin⁡θ,Y=−xsin⁡θ+ycos⁡θ.\begin{aligned} X &= x\cos\theta + y\sin\theta, \\[0.4em] Y &= -x\sin\theta + y\cos\theta. \end{aligned}

The first pair is the rotation through θ\theta applied to the point (X,Y)(X, Y), because the new axes are the old axes rotated through θ\theta. The second pair is the rotation through −θ-\theta.

In plain words The points stay put and the grid turns. Turning the grid through θ\theta changes the coordinates the same way as turning the points through −θ-\theta.

Example Take θ=π4\theta = \frac{\pi}{4}, so cos⁡θ=sin⁡θ=12\cos\theta = \sin\theta = \frac{1}{\sqrt{2}}. Then

x=X−Y2,y=X+Y2,xy=X2−Y22.\begin{aligned} x &= \frac{X - Y}{\sqrt{2}}, \\[0.8em] y &= \frac{X + Y}{\sqrt{2}}, \\[0.8em] xy &= \frac{X^2 - Y^2}{2}. \end{aligned}

So the curve xy=2xy = 2 has the equation X2−Y2=4X^2 - Y^2 = 4 in the new coordinates, which is the hyperbola

X24−Y24=1\frac{X^2}{4} - \frac{Y^2}{4} = 1

with a=b=2a = b = 2 and c=22c = 2\sqrt{2}. Its vertices (X,Y)=(±2,0)(X, Y) = (\pm 2, 0) are the points (2,2)(\sqrt{2}, \sqrt{2}) and (−2,−2)(-\sqrt{2}, -\sqrt{2}) in the old coordinates, and its foci are (2,2)(2, 2) and (−2,−2)(-2, -2). As a check, 2⋅2=2\sqrt{2} \cdot \sqrt{2} = 2.

Mixed xy term

Definition In a second degree equation

Ax2+Bxy+Cy2+Dx+Ey+F=0,\begin{aligned} &Ax^2 + Bxy + Cy^2 \\[0.4em] &\quad + Dx + Ey + F = 0, \end{aligned}

with AA, BB and CC not all zero, the term BxyBxy is called the mixed term or cross term. Rotating the axes through θ\theta turns the equation into one of the same kind in XX and YY, with the coefficients

A′=Acos⁡2θ+Bsin⁡θcos⁡θ+Csin⁡2θ,B′=Bcos⁡2θ−(A−C)sin⁡2θ,C′=Asin⁡2θ−Bsin⁡θcos⁡θ+Ccos⁡2θ,D′=Dcos⁡θ+Esin⁡θ,E′=−Dsin⁡θ+Ecos⁡θ,F′=F.\begin{aligned} A' &= A\cos^2\theta + B\sin\theta\cos\theta \\ &\quad + C\sin^2\theta, \\[0.4em] B' &= B\cos 2\theta - (A - C)\sin 2\theta, \\[0.4em] C' &= A\sin^2\theta - B\sin\theta\cos\theta \\ &\quad + C\cos^2\theta, \\[0.4em] D' &= D\cos\theta + E\sin\theta, \\[0.4em] E' &= -D\sin\theta + E\cos\theta, \\[0.4em] F' &= F. \end{aligned}

For a parabola, a hyperbola or an ellipse that is not a circle, the mixed term is nonzero exactly when the axes of the conic are not parallel to the coordinate axes.

In plain words An xyxy term is the sign of a tilted curve. Tilting a conic other than a circle creates one, and rotating the axes by the right angle removes it.

Example Rotate the ellipse x24+y2=1\frac{x^2}{4} + y^2 = 1 through π4\frac{\pi}{4}. By the rule for rotating a curve, replace xx by x+y2\frac{x + y}{\sqrt{2}} and yy by y−x2\frac{y - x}{\sqrt{2}}, which gives

(x+y)28+(y−x)22=1.\frac{(x + y)^2}{8} + \frac{(y - x)^2}{2} = 1.

Multiplying by 88 and expanding gives 5x2−6xy+5y2=85x^2 - 6xy + 5y^2 = 8. The mixed term −6xy-6xy appeared because the major axis now lies along the line y=xy = x. The vertex (2,0)(2, 0) has moved to (2,2)(\sqrt{2}, \sqrt{2}), and indeed 5⋅2−6⋅2+5⋅2=85 \cdot 2 - 6 \cdot 2 + 5 \cdot 2 = 8.

Angle convention

Definition The rotation angle θ\theta is measured from the positive xx-axis to the positive XX-axis, counterclockwise when positive, and it may be given in radians or degrees. The mixed term disappears when B′=0B' = 0, that is when

Bcos⁡2θ=(A−C)sin⁡2θ.B\cos 2\theta = (A - C)\sin 2\theta.

When B=0B = 0 there is nothing to remove and no rotation is needed. When B≠0B \ne 0 the standard choice is the unique angle with 0<θ<π20 < \theta < \frac{\pi}{2} and

cot⁡2θ=A−CB.\cot 2\theta = \frac{A - C}{B}.

Since 0<2θ<π0 < 2\theta < \pi, the value sin⁡2θ\sin 2\theta is positive and cos⁡2θ\cos 2\theta has the sign of A−CB\frac{A - C}{B}. So θ<π4\theta < \frac{\pi}{4} when that fraction is positive, θ=π4\theta = \frac{\pi}{4} when A=CA = C, and θ>π4\theta > \frac{\pi}{4} when it is negative.

In plain words Turn the axes counterclockwise by less than a right angle. Adding or removing a quarter turn would also clear the mixed term, but it only swaps the roles of XX and YY, so the convention fixes one answer.

Example An equation with A=3A = 3, B=23B = 2\sqrt{3} and C=1C = 1 has cot⁡2θ=223=13\cot 2\theta = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}, so 2θ=π32\theta = \frac{\pi}{3} and θ=π6\theta = \frac{\pi}{6}. An equation with A=1A = 1, B=23B = 2\sqrt{3} and C=3C = 3 has cot⁡2θ=−13\cot 2\theta = -\frac{1}{\sqrt{3}}, so 2θ=2π32\theta = \frac{2\pi}{3} and θ=π3\theta = \frac{\pi}{3}. A calculator would turn tan⁡2θ=−3\tan 2\theta = -\sqrt{3} into 2θ=−π32\theta = -\frac{\pi}{3}, which is outside the convention. Adding π\pi gives the correct 2θ=2π32\theta = \frac{2\pi}{3}.

Eliminating the mixed term

Definition To remove the mixed term from an equation with B≠0B \ne 0, first find the angle θ\theta of the convention. When it is not a familiar angle, compute

cos⁡2θ=cot⁡2θ1+cot⁡22θ,\cos 2\theta = \frac{\cot 2\theta}{\sqrt{1 + \cot^2 2\theta}},

which holds because sin⁡2θ>0\sin 2\theta > 0. The half-angle formulas, with positive roots because 0<θ<π20 < \theta < \frac{\pi}{2}, then give

cos⁡θ=1+cos⁡2θ2,sin⁡θ=1−cos⁡2θ2.\begin{aligned} \cos\theta &= \sqrt{\frac{1 + \cos 2\theta}{2}}, \\[0.8em] \sin\theta &= \sqrt{\frac{1 - \cos 2\theta}{2}}. \end{aligned}

Finally substitute x=Xcos⁡θ−Ysin⁡θx = X\cos\theta - Y\sin\theta and y=Xsin⁡θ+Ycos⁡θy = X\sin\theta + Y\cos\theta, or use the coefficient formulas. The result has no XYXY term, and completing the square finishes the job.

In plain words Find the angle, find its cosine and sine, substitute, and simplify. You never need the angle itself, only its cosine and sine.

Example Take 32x2−72xy+53y2=8032x^2 - 72xy + 53y^2 = 80. Here

cot⁡2θ=32−53−72=724,\cot 2\theta = \frac{32 - 53}{-72} = \frac{7}{24},

so cos⁡2θ=725\cos 2\theta = \frac{7}{25}, which gives cos⁡θ=45\cos\theta = \frac{4}{5} and sin⁡θ=35\sin\theta = \frac{3}{5}, an angle of about 36.87∘36.87^\circ. The coefficient formulas give

A′=32⋅16−72⋅12+53⋅925=5,C′=32⋅9+72⋅12+53⋅1625=80.\begin{aligned} A' &= \frac{32 \cdot 16 - 72 \cdot 12 + 53 \cdot 9}{25} \\[0.4em] &= 5, \\[0.8em] C' &= \frac{32 \cdot 9 + 72 \cdot 12 + 53 \cdot 16}{25} \\[0.4em] &= 80. \end{aligned}

The equation becomes 5X2+80Y2=805X^2 + 80Y^2 = 80, that is

X216+Y2=1.\frac{X^2}{16} + Y^2 = 1.

It is an ellipse with a=4a = 4 and b=1b = 1. Its major axis lies along the line y=34xy = \frac{3}{4}x, and the vertex (X,Y)=(4,0)(X, Y) = (4, 0) is the point (165,125)(\frac{16}{5}, \frac{12}{5}), which satisfies the original equation.

Classification by the discriminant

Definition The discriminant of the equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 is the number B2−4ACB^2 - 4AC. It does not change under a rotation of axes,

(B′)2−4A′C′=B2−4AC.(B')^2 - 4A'C' = B^2 - 4AC.

After a rotation that removes the mixed term it equals −4A′C′-4A'C', so the test for shifted conics gives the following.

  • If B2−4AC<0B^2 - 4AC < 0, the graph is an ellipse (possibly a circle), a single point, or empty.
  • If B2−4AC=0B^2 - 4AC = 0, the graph is a parabola, two parallel lines, one line, or empty.
  • If B2−4AC>0B^2 - 4AC > 0, the graph is a hyperbola or two crossing lines.

So when the graph is known to be a nondegenerate conic, the sign names its type. The sign alone cannot tell a true conic from a degenerate one. For that, rotate and complete the square, or factor.

In plain words Compute B2−4ACB^2 - 4AC first. Negative points to an ellipse, zero to a parabola and positive to a hyperbola. But the curve might have collapsed into lines or a point, or might not exist at all, so the sign says what the graph can be, not what it must be.

Example For 32x2−72xy+53y2=8032x^2 - 72xy + 53y^2 = 80 the discriminant is 5184−6784=−1600<05184 - 6784 = -1600 < 0, and the curve is a true ellipse, as found above. The equations x2+2xy+y2+x−y=0x^2 + 2xy + y^2 + x - y = 0 and x2+2xy+y2−1=0x^2 + 2xy + y^2 - 1 = 0 both have discriminant 4−4=04 - 4 = 0. Rotating the first through π4\frac{\pi}{4} gives 2X2−2 Y=02X^2 - \sqrt{2}\,Y = 0, a parabola. The second is (x+y)2=1(x + y)^2 = 1, the pair of parallel lines x+y=1x + y = 1 and x+y=−1x + y = -1. Likewise x2−xy−2y2=0x^2 - xy - 2y^2 = 0 has the positive discriminant 1+8=91 + 8 = 9, but it factors as (x−2y)(x+y)=0(x - 2y)(x + y) = 0 and describes two lines through the origin.

Check yourself

  1. Rotate the axes through π4\frac{\pi}{4}. What equation does xy=−8xy = -8 become, and where are the vertices of the curve?
Show the answer

Since xy=X2−Y22xy = \frac{X^2 - Y^2}{2}, the equation becomes X2−Y2=−16X^2 - Y^2 = -16, that is

Y216−X216=1.\frac{Y^2}{16} - \frac{X^2}{16} = 1.

It is a hyperbola whose transverse axis lies along the YY-axis, the line y=−xy = -x. Its vertices (X,Y)=(0,±4)(X, Y) = (0, \pm 4) are the points (−22,22)(-2\sqrt{2}, 2\sqrt{2}) and (22,−22)(2\sqrt{2}, -2\sqrt{2}), and each has xy=−8xy = -8.

  1. Find the angle θ\theta with 0<θ<π20 < \theta < \frac{\pi}{2} that removes the mixed term from 7x2−63 xy+13y2=167x^2 - 6\sqrt{3}\,xy + 13y^2 = 16, and write the new equation.
Show the answer

The angle satisfies

cot⁡2θ=7−13−63=13,\cot 2\theta = \frac{7 - 13}{-6\sqrt{3}} = \frac{1}{\sqrt{3}},

so 2θ=π32\theta = \frac{\pi}{3} and θ=π6\theta = \frac{\pi}{6}. The coefficient formulas give A′=4A' = 4 and C′=16C' = 16, so the equation becomes 4X2+16Y2=164X^2 + 16Y^2 = 16, which is the ellipse

X24+Y2=1.\frac{X^2}{4} + Y^2 = 1.
  1. The equation x2−4xy+4y2=9x^2 - 4xy + 4y^2 = 9 has discriminant 00. Is its graph a parabola?
Show the answer

No. The left side is (x−2y)2(x - 2y)^2, so the equation says x−2y=3x - 2y = 3 or x−2y=−3x - 2y = -3. The graph is a pair of parallel lines, one of the degenerate cases.

How this lesson connects

Where this leads in Calculus