Rotation of Axes
Rotation of Coordinate Systems
It is often convenient to analyze curves and equations in terms of a Cartesian coordinate system for which the axes have been rigidly rotated through a (normally acute) angle with respect to the standard Cartesian coordinate system.
Transformation of Coordinates Under Rotation
Let P be a point in the plane; then P has coordinates $ (x,y) $ in the standard Cartesian coordinate system (called the old system) and coordinates $ (x’,y’) $ in the rotated system (called the new system). (See Fig. 39-1.) Then the coordinates in the old system can be expressed in terms of the coordinates in the new system by the transformation equations:
\[ \begin{aligned}&x=x^{\prime}\cos\theta-y^{\prime}\sin\theta\\&y=x^{\prime}\sin\theta+y^{\prime}\cos\theta\\ \end{aligned} \]
These equations can be applied to the coordinates of individual points; a frequent use is to transform equations of curves given in the old coordinate system into equations in the new system, where the form of the equation may be easier to analyze.
EXAMPLE 39.1 Analyze the effect on the equation xy = 2 of rotating the axes through a $ 45^{} $ angle.
If $ = 45^{} $ , then $ = = 1/ $ . Hence the transformation equations become
\[ x=\frac{x^{\prime}-y^{\prime}}{\sqrt{2}},\quad y=\frac{x^{\prime}+y^{\prime}}{\sqrt{2}} \]
Performing these substitution in the original equation yields
\[ xy=\left(\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}\right)\left(\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}\right)=2 \]
\[ \frac{x^{\prime2}-y^{\prime2}}{2}=2 \]
This can be written as
\[ \frac{x^{\prime2}}{4}-\frac{y^{\prime2}}{4}=1 \]
which can be seen to be the equation of a hyperbola in standard position with foci on the $ x^{} $ -axis (that is, the new x-axis), rotated $ 45^{} $ with respect to the old.
Analyzing Second-Degree Equations
In analyzing second-degree equations written in the standard form
\[ Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0 \]
it is useful to rotate axes. An angle $ $ can always be found such that rotating axes through this angle transforms the equation into the form $ A{}x{} + C{}y{} + D{}x{} + E{}y{} + F = 0 $ . The angle $ $ is given by
If A = C, then $ = 45^{} $
Otherwise, $ $ is a solution of the equation $ = $ .
SOLVED PROBLEMS
39.1. Show that for any point P that has coordinates $ (x,y) $ in a standard Cartesian coordinate system and coordinates $ (x’,y’) $ in a system with axes rotated through $ $ , the transformation equations $ x = x’- y’$ , $ y = x’+ y’$ hold.
Consider the vector $ $ drawn from the origin of both coordinate systems to P in Fig. 39-2.
It is convenient to use the notation of Problem 27.13, in which i and j are defined, respectively, as the unit vectors in the positive x- and y-directions (in the old coordinate system). Then, in this coordinate system, $ = x + y $ . Similarly, $ i’ $ and $ j’ $ are, respectively, the unit vectors in the positive $ x’ $ - and $ y’ $ -directions, and, in this coordinate system, $ = x’i’ + y’j’ $ . Then $ x + y = x’i’ + y’j’ $ . If, now, the dot product of both sides of this identity is taken with vector i, it follows that:
\[ (x\mathbf{i}+y\mathbf{j})\cdot\mathbf{i}=(x^{\prime}\mathbf{i}^{\prime}+y^{\prime}\mathbf{j}^{\prime})\cdot\mathbf{i} \]
\[ x\mathbf{i}\cdot\mathbf{i}+y\mathbf{j}\cdot\mathbf{i}=x^{\prime}\mathbf{i}^{\prime}\cdot\mathbf{i}+y^{\prime}\mathbf{j}^{\prime}\cdot\mathbf{i} \]
In the latter identity, apply the theorem on the dot product:
Since the angle between i and i is $ 0^{} $ , $ i i = lil ^{} = 1 = 1 $ .
Since the angle between i and j is $ 90^{} $ , $ i j = |i||j|^{} = 1 = 0 $ .
Since the angle between i and $ i’ $ is $ $ , $ i i’ = |i||i’| = 1 = $ .
Since the angle between i and $ j’ $ is $ + 90^{} $ , $ i j’ = |i||j’|(+ 90^{}) = 1 (-) = -$ .
Substituting yields:
\[ x(1)+y(0)=x^{\prime}\cos\theta-y^{\prime}\sin\theta \]
\[ x=x^{\prime}\cos\theta-y^{\prime}\sin\theta \]
The proof of the transformation equation for y, $ y = x’ + y’ $ , is left as an exercise.
39.2. Show that an angle $ $ can always be found such that rotating axes through this angle transforms the equation $ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 $ into the equation $ A’x’^{2} + C’y’^{2} + D’x’ + E’y’ + F = 0 $ .
Rotating axes through an angle \(\theta\) transforms the equation \(Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0\) by making the substitutions \(x = x^{\prime}\cos\theta - y^{\prime}\sin\theta\), \(y = x^{\prime}\sin\theta + y^{\prime}\cos\theta\). Performing the substitutions yields:
\[ A(x^{\prime}\cos\theta-y^{\prime}\sin\theta)^{2}+B(x^{\prime}\cos\theta-y^{\prime}\sin\theta)(x^{\prime}\sin\theta+y^{\prime}\cos\theta)+C(x^{\prime}\sin\theta+y^{\prime}\cos\theta)^{2} \]
\[ +D(x^{\prime}\cos\theta-y^{\prime}\sin\theta)+E(x^{\prime}\sin\theta+y^{\prime}\cos\theta)+F=0 \]
Expanding and combining terms in $ x^{} $ , $ y^{} $ , $ x{}y{} $ , $ x^{} $ , and $ y^{} $ yields:
\[ \begin{aligned}&x^{\prime2}(A\cos^{2}\theta+B\cos\theta\sin\theta+C\sin^{2}\theta)+x^{\prime}y^{\prime}[-2A\cos\theta\sin\theta+B(\cos^{2}\theta-\sin^{2}\theta)+2C\sin\theta\cos\theta]\\&\quad+y^{\prime2}(A\sin^{2}\theta-B\sin\theta\cos\theta+C\cos^{2}\theta)+x^{\prime}(D\cos\theta+E\sin\theta)+y^{\prime}(-D\sin\theta+E\cos\theta)+F=0\\ \end{aligned} \]
In order for the equation to have exactly the form $ A{}x{} + C{}y{} + D{}x{} + E{}y{} + F = 0 $ , the coefficient of the $ x{}y{} $ term must be zero, that is:
\[ \begin{aligned}-2A\cos\theta\sin\theta+B(\cos^{2}\theta-\sin^{2}\theta)+2C\sin\theta\cos\theta&=0\\-A\sin2\theta+B\cos2\theta+C\sin2\theta&=0\ $ A-C)\sin2\theta&=B\cos2\theta\end{aligned} \]
Thus if A = C, then $ B = 0 $ , thus $ 2= 90^{} $ , or $ = 45^{} $ . Otherwise, divide both sides by $ (A - C) $ to obtain
\[ \frac{\sin2\theta}{\cos2\theta}=\frac{B}{A-C} \]
\[ \tan2\theta=\frac{B}{A-C} \]
This equation will have an acute angle solution for $ $ .
39.3. Find an appropriate angle through which to rotate axes and sketch a graph of the equation
\[ 3x^{2}-2\sqrt{3}xy+y^{2}+2x+2\sqrt{3}y=0. \]
Here A = 3, $ B = -2 $ , C = 1; hence set
\[ \tan2\theta=\frac{B}{A-C}=\frac{-2\sqrt{3}}{3-1}=-\sqrt{3} \]
The smallest solution of this equation is given by $ 2= 120^{} $ , that is, $ = 60^{} $ . Since $ ^{} = /2 $ and $ ^{} = $ , the transformation equations are
\[ x=\frac{x^{\prime}-y^{\prime}\sqrt{3}}{2}\qquad y=\frac{x^{\prime}\sqrt{3}+y^{\prime}}{2} \]
Substituting these into the original equation followed by simplification yields:
\[ \begin{aligned}3\Big(\frac{x^{\prime}-y^{\prime}\sqrt{3}}{2}\Big)^{2}-2\sqrt{3}\Big(\frac{x^{\prime}-y^{\prime}\sqrt{3}}{2}\Big)\Big(\frac{x^{\prime}\sqrt{3}+y^{\prime}}{2}\Big)+\Big(\frac{x^{\prime}\sqrt{3}+y^{\prime}}{2}\Big)^{2}+2\Big(\frac{x^{\prime}-y^{\prime}\sqrt{3}}{2}\Big)\\+2\sqrt{3}\Big(\frac{x^{\prime}\sqrt{3}+y^{\prime}}{2}\Big)=0\\\frac{x^{\prime2}(3-6+3)+x^{\prime}y^{\prime}(-6\sqrt{3}+4\sqrt{3}+2\sqrt{3})+y^{\prime2}(9+6+1)}{4}+\frac{x^{\prime}(2+6)+y^{\prime}(-2\sqrt{3}+2\sqrt{3})}{2}=0\\4y^{\prime2}+4x^{\prime}=0\\y^{\prime2}=-x^{\prime}\end{aligned} \]
Thus, in the rotated system, the graph of the equation is a parabola, vertex at $ (0,0) $ , opening left. The graph is shown in Fig. 39-3.
39.4. Solve the transformation equations for $ x’ $ and $ y’ $ in terms of x and y to find the reverse transformation equations.
Write the transformation equations in standard form for equations in $ x’ $ and $ y’ $ :
\[ \begin{aligned}&x^{\prime}\cos\theta-y^{\prime}\sin\theta=x\\&x^{\prime}\sin\theta+y^{\prime}\cos\theta=y\\ \end{aligned} \]
Now apply Cramer’s rule to obtain:
\[ x^{\prime}=\frac{\left|\begin{array}{cc}{{{x}}}&{{{-\sin\theta}}} \\{{{y}}}&{{{\cos\theta}}}\end{array}\right|}{\left|\begin{array}{cc}{{{\cos\theta}}}&{{{-\sin\theta}}} \\{{{\sin\theta}}}&{{{\cos\theta}}}\end{array}\right|}=x\cos\theta+y\sin\theta\qquad y^{\prime}=\frac{\left|\begin{array}{cc}{{{\cos\theta}}}&{{{x}}} \\{{{\sin\theta}}}&{{{y}}}\end{array}\right|}{\left|\begin{array}{cc}{{{\cos\theta}}}&{{{-\sin\theta}}} \\{{{\sin\theta}}}&{{{\cos\theta}}}\end{array}\right|}=-x\sin\theta+y\cos\theta \]
39.5. Find appropriate transformation equations for rotation of axes and sketch a graph of the equation $ 2x{2}-3xy-2y{2}+10=0 $ .
Here A = 2, B = -3, C = -2; hence set
\[ \tan2\theta=\frac{B}{A-C}=\frac{-3}{2-(-2)}=-\frac{3}{4} \]
An exact solution of this equation is not possible, but it is also not necessary, since $ $ and $ $ can be found from the half-angle formulas. Assume the smallest solution of the equation, with $ 90{}<2<180{} $ .
then, since $ =-=-=-,==-. $
Hence, since $ 45^{} < < 90^{} $ ,
\[ \sin\theta=\sqrt{\frac{1-\cos2\theta}{2}}=\sqrt{\frac{1-(-4/5)}{2}}=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}} \]
\[ \cos\theta=\sqrt{\frac{1+\cos2\theta}{2}}=\sqrt{\frac{1+(-4/5)}{2}}=\sqrt{\frac{1}{10}}=\frac{1}{\sqrt{10}} \]
Thus the transformation equations to rotate axes in order to eliminate the xy term are:
\[ x=\frac{x^{\prime}-3y^{\prime}}{\sqrt{10}}\qquad y=\frac{3x^{\prime}+y^{\prime}}{\sqrt{10}} \]
Substituting these into the original equation followed by simplification yields:
\[ 2\bigg(\frac{x^{\prime}-3y^{\prime}}{\sqrt{10}}\bigg)^{2}-3\bigg(\frac{x^{\prime}-3y^{\prime}}{\sqrt{10}}\bigg)\bigg(\frac{3x^{\prime}+y^{\prime}}{\sqrt{10}}\bigg)-2\bigg(\frac{3x^{\prime}+y^{\prime}}{\sqrt{10}}\bigg)^{2}+10=0 \]
\[ \begin{aligned}\frac{x^{\prime2}(2-9-18)+x^{\prime}y^{\prime}(-12+24-12)+y^{\prime2}(18+9-2)}{10}+10=0\\\frac{-25x^{\prime2}+25y^{\prime2}}{10}+10=0\\\frac{x^{\prime2}}{4}-\frac{y^{\prime2}}{4}=1\end{aligned} \]
Thus, in the rotated system, the graph of the equation is a hyperbola in standard position, with focal axis on the $ x^{} $ axis and asymptotes $ y^{} = x^{} $ . To sketch, note that the axes have been rotated through an angle $ $ with $ = 3 $ ; hence the $ x^{} $ -axis has precisely slope 3 with respect to the old coordinate system. The graph, together with the asymptotes, is shown in Fig. 39-4.
39.6. In the previous problem, (a) find the coordinates of the foci in the new system and in the old system and (b) find the equations of the asymptotes in the old system.
- From the equation of the hyperbola in the new system, a = b = 2; hence $ c = = 2 $ .
Thus the coordinates of the foci in the new system are \((x', y') = (\pm 2\sqrt{2}, 0)\). To transform these to the old system, use the transformation equations:
\[ x=x^{\prime}\cos\theta-y^{\prime}\sin\theta=\pm2\sqrt{2}\bigg(\frac{1}{\sqrt{10}}\bigg)-0\bigg(\frac{3}{\sqrt{10}}\bigg)=\pm\frac{2}{\sqrt{5}} \]
\[ y=x^{\prime}\sin\theta+y^{\prime}\cos\theta=\pm2\sqrt{2}\bigg(\frac{3}{\sqrt{10}}\bigg)+0\bigg(\frac{1}{\sqrt{10}}\bigg)=\pm\frac{6}{\sqrt{5}} \]
Hence the coordinates of the foci in the old system are $ (x, y) = ( , ) $ and $ (x, y) = ( -, - ) $ .
- The equations of the asymptotes in the new system are $ y’ = x’ $ . To transform these to the old system, use the reverse transformation equations:
\[ x^{\prime}=x\cos\theta+y\sin\theta=\frac{x+3y}{\sqrt{10}}\qquad y^{\prime}=-x\sin\theta+y\cos\theta=\frac{-3x+y}{\sqrt{10}} \]
Then $ y’ = x’ $ becomes $ = $ , or, after simplification, -2x = y, and $ y’ = -x’ $ becomes $ = -() $ , or, after simplification, x = 2y.
SUPPLEMENTARY PROBLEMS
39.7. Complete Problem 39.1 by showing that a rotation through an angle $ $ transforms y according to the transformation equation $ y = x’ + y’ $ .
39.8. Find an appropriate angle through which to rotate axes to eliminate the xy term in the equation $ 21x^{2} - 10xy + 31y^{2} = 144 $ .
Ans. $ 30^{} $
39.9. Find the equation into which $ 21x^{2} - 10xy + 31y^{2} = 144 $ is transformed by the rotation of the previous problems and sketch the graph.
Ans. Equation: $ +=1 $ . See Fig. 39-5.
39.10. Show that the equation $ x^{2} + y^{2} = r^{2} $ does not change (is invariant) under a rotation of the axes through any angle $ $ .
39.11. Find the transformation equations to rotate axes through an appropriate angle to eliminate the xy term in the equation $ 16x^{2} + 24xy + 9y^{2} + 60x - 80y + 100 = 0 $ .
Ans. $ x = $ , $ y = $
39.12. Find the equation into which $ 16x^{2} + 24xy + 9y^{2} + 60x - 80y + 100 = 0 $ is transformed by the rotation of the previous problem and sketch the graph.
Ans. Equation: $ x’^{2} = 4(y’ - 1) $ . See Fig. 39-6.
39.13. Show that in transforming the equation $ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 $ through any rotation of axes into an equation of form $ A’x’^{2} + B’x’y’ + C’y’^{2} + D’x’ + E’y’ + F = 0 $ , the quantity $ A + C $ will equal the quantity $ A’ + C’ $ . (Hint: See Problem 39.2 for expressions for $ A’ $ and $ C’ $ .)
39.14. Find the transformation equations to rotate axes through an appropriate angle to eliminate the xy term in the equation $ 3x^{2} + 8xy - 3y^{2} - 4x + 8y = 0 $ .
\[ x=\frac{2x^{\prime}-y^{\prime}}{\sqrt{5}},y=\frac{x^{\prime}+2y^{\prime}}{\sqrt{5}}. \]
39.15. Find the equation into which $ 3x^{2} + 8xy - 3y^{2} - 4x + 8y = 0 $ is transformed by the rotation of the previous problem and sketch the graph.
Ans. Equation: $ - = 1 $ . See Fig. 39-7.
