Rotation of Axes 1. Rotation of Axes. At the beginning of Chapter 5 we stated that all equations of the form

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1 Rotation of Axes 1 Rotation of Axes At the beginning of Chapter we stated that all equations of the form Ax + Bx + C + Dx + E + F =0 represented a conic section, which might possibl be degenerate. We saw in Section. that the graph of the quadratic equation Ax + C + Dx + E + F =0 is a parabola when A =0orC = 0, that is, when AC = 0. In Section.3 we found that the graph is an ellipse if AC > 0, and in Section.4 we saw that the graph is a hperbola when AC < 0. So we have classified the situation when B = 0. Degenerate situations can occur; for example, the quadratic equation x = 0 has no solutions, and the graph of x = 0 is not a hperbola, but the pair of lines with equations = ±x. But excepting this tpe of situation, we have categorized the graphs of all the quadratic equations with B =0. When B 0, a rotation can be performed to bring the conic into a form that will allow us to compare it with one that is in standard position. To set up the discussion, letusfirstsupposethatasetofx-coordinate axes has been rotated about the origin b an angle θ, where 0 <θ<π/, to form a new set of ˆx-axes, as shown in Figure 1(a). We would like to determine the coordinates for a point P in the plane relative to the two coordinate sstems. P(x, ) P(x, ˆ ) ˆ P(x, ) P(x, ˆ ) ˆ r B r f B u A x u A x (a) (b) Figure 1

3 we found that the graph is an ellipse if AC > 0, and in Section.4 we saw that the graph is a hperbola when AC < 0. So we have classified the situation when B = 0.

2 CHAPTER Conic Sections, Polar Coordinates, and Parametric Equations First introduce a new pair of variables r and φ to represent, respectivel, the distance from P to the origin and the angle formed b the ˆx-axis and the line connecting the origin to P, as shown in Figure 1(b). From the right triangle OBP shown in Figure (a) we see that ˆx = r cos φ and = r sin φ, and from the right triangle OAP shown in Figure (b) we see that x = r cos(φ + θ) and = r sin(φ + θ). P(x, ˆ ) ˆ P(x, ) f r B r O O f 1 u x A (a) (b) Figure we have Using the fundamental trigonometric identities for the sum of the sine and the cosine x = r cos(φ + θ) =r cos φ cos θ r sin φ sin θ and = r sin(φ + θ) =r cos φ sin θ + r sin φ cos θ. Since ˆx = r cos φ and = r sin φ, we have the following result. The second pair of equations is derived b solving for ˆx and in the first pair.

From the right triangle OBP shown in Figure (a) we see that ˆx = r cos φ and = r sin φ, and from the right triangle OAP shown in Figure (b) we see that x = r cos(φ + θ) and = r sin(φ + θ).

3 Rotation of Axes 3 Coordinate Rotation Formulas If a rectangular x-coordinate sstem is rotated through an angle θ to form an ˆxcoordinate sstem, then a point P (x, ) will have coordinates P (ˆx, ) in the new sstem, where (x, ) and(ˆx, ) are related b and x =ˆx cos θ sin θ and =ˆx sin θ + cos θ. ˆx = x cos θ + sin θ and = x sin θ + cos θ. EXAMPLE 1 Show that the graph of the equation x = 1 is a hperbola b rotating the x-axes through an angle of π/4. Solution: Denoting a point in the rotated sstem b (ˆx, ), we have and x =ˆx cos π 4 sin π 4 = (ˆx ) =ˆx sin π 4 + cos π 4 = (ˆx +). Substituting these expressions into the original equation x = 1 produces the equation (ˆx ) (ˆx +) =1 or ˆx =. In the ˆx-coordinate sstem, then, we have a standard position hperbola whose asmptotes are = ±ˆx. These are the same lines as the x- and-axes, as seen in Figure 3.

EXAMPLE 1 Show that the graph of the equation x = 1 is a hperbola b rotating the x-axes through an angle of π/4.

4 4 CHAPTER Conic Sections, Polar Coordinates, and Parametric Equations x 1 ˆ d x Figure 3 In Example 1 the appropriate angle of rotation was provided to eliminate the ˆx-term from the equation. Such an angle θ can alwas be found so that when the coordinate axes are rotated through this angle, the equation in the new coordinate sstem will not involve the product ˆx. To determine the angle, suppose we have the general second-degree equation Ax + Bx + C + Dx + E + F =0, where B 0. Introduce the variables x =ˆx cos θ sin θ and =ˆx sin θ + cos θ, and substitute for x and in the original equation. This gives us the new equation in ˆx and : A(ˆx cos θ sin θ) + B(ˆx cos θ sin θ)(ˆx sin θ + cos θ)+c(ˆx sin θ + cos θ) + D(ˆx cos θ sin θ)+e(ˆx sin θ + cos θ)+f =0. Performing the multiplication and collecting the similar terms gives ˆx ( A(cos θ) + B(cos θ sin θ)+c(sin θ) ) +ˆx [ A cos θ sin θ + B ( (cos θ) (sin θ) ) +C sin θ cos θ ] + ( A(sin θ) B(sin θ cos θ)+c(cos θ) ) +ˆx(D cos θ + E sin θ)+( D sin θ + E cos θ)+f =0.

To determine the angle, suppose we have the general second-degree equation Ax + Bx + C + Dx + E + F =0, where B 0.

5 Rotation of Axes To eliminate the ˆx-term from this equation, choose θ so that the coefficient of this term is zero, that is, so that A cos θ sin θ + B ( (cos θ) (sin θ) ) +C sin θ cos θ =0. Simplifing this equation, we have and B ( (cos θ) (sin θ) ) =(A C)cosθ sin θ (cos θ) (sin θ) cosθ sin θ = A C B. Using the double angle formulas for sine and cosine gives cos θ sin θ = A C B, or cot θ = A C B. Rotating Quadratic Equations To eliminate the x-term in the general quadratic equation Ax + Bx + C + Dx + E + F =0, where B 0, rotate the coordinate axes through an angle θ that satisfies cot θ = A C B. EXAMPLE Sketch the graph of 4x 4x +7 4 = 0. Solution: To eliminate the x-term we first rotate the coordinate axes through the angle θ where cot θ = A C B = = 3 4 From the triangle in Figure 4, we see that cos θ = 3.

Using the double angle formulas for sine and cosine gives cos θ sin θ = A C B, or cot θ = A C B.

6 6 CHAPTER Conic Sections, Polar Coordinates, and Parametric Equations Using the half angle formulas we have 1+ 3 cos θ = =, and sin θ = 1 3 =. cot u! cos u E 4 u 3 Figure 4 The rotation of axes formulas then give x = ˆx and = ˆx +. Substituting these into the original equation gives ( ) ( 4 ˆx )( 4 ˆx ˆx + ) ( +7 ˆx + ) 4 = 0. After simplifing we have ˆx +8 =4 or ˆx =1. This is the graph of an ellipse in standard position with respect to the ˆx coordinate sstem. The graph of the equation 4x 4x +7 =4 is consequentl the graph of an ellipse that has been rotated through an angle θ = arcsin 6, asshowninfigure.

Substituting these into the original equation gives ( ) ( 4 ˆx )( 4 ˆx ˆx + ) ( +7 ˆx + ) 4 = 0. After simplifing we have ˆx +8 =4 or ˆx 10 + 3 =1.

7 Rotation of Axes 7 4 u arcsin(œ/) 4 4 x 4x 4x Figure There is an easil applied formula that can be used to determine which conic will be produced once the rotation has been performed. Suppose that a rotation θ of the coordinate axes changes the equation Ax + Bx + C + Dx + E + F =0 into the equation Âˆx + ˆBˆx + Ĉ + ˆDˆx + Ê + ˆF =0. It is not hard to show, although it is algebraicall tedious, that the coefficients of these two equations satisf B 4AC = ˆB 4ÂĈ. When the particular angle chosen is such that ˆB =0,wehave B 4AC = 4ÂĈ. However, except for the degenerate cases we know that the new equation Âˆx + Ĉ + ˆDˆx + Ê + ˆF =0 will be: i) An ellipse if ÂĈ >0; ii) A hperbola if ÂĈ <0; iii) A parabola if ÂĈ =0. Appling this result to the original equation gives the following.

It is not hard to show, although it is algebraicall tedious, that the coefficients of these two equations satisf B 4AC = ˆB 4ÂĈ.

8 8 CHAPTER Conic Sections, Polar Coordinates, and Parametric Equations Classification of Conic Curves Except for degenerate cases, the equation Ax + Bx + C + Dx + E + F =0 will be: i) An ellipse if B 4AC = 4ÂĈ <0; ii) A hperbola if B 4AC = 4ÂĈ >0; iii) A parabola if B 4AC = ÂĈ =0. The beaut of this result is that we do not need to know the angle that produces the elimination of the ˆx-terms. We onl need to know that such an angle exists, and we have alread established this fact. If we appl the result to the examples in this section we see that: Example 1. The equation x =1has B 4AC =1 4(0)(0) = 1 > 0 and is therefore a hperbola. Example. The equation 4x 4x +7 4 = 0 has B 4AC =( 4) 4(4)(7) = 6 < 0 and is therefore an ellipse. We strongl recommend that ou appl this simple test when rotation is required to graph a general quadratic equation. It provides a final check on a result produced from a considerable amount of algebraic and trigonometric computation. While it will not ensure that ou are correct, it can tell ou when ou are certainl wrong, and this is often more important.

We onl need to know that such an angle exists, and we have alread established this fact. If we appl the result to the examples in this section we see that: Example 1.

9 Rotation of Axes 9 Rotation of Axes Exercises In Exercises 1 4, determine the ˆx-coordinates of the given point if the coordinate axes are rotated through the given angle θ. 1. (0, 1), θ=60. (1, 1), θ=4 3. ( 3, 1), θ=30 4. (, ), θ=90 In Exercises 1, (a) determine whether the conic section is an ellipse, hperbola, or parabola, and (b) perform a rotation, and if necessar a translation, and sketch the graph.. x x + = 6. x +4x + = x 6x +9 =0 8. x 3x =1 9. 4x 4x +7 =4 10. x 3x + = 11. 6x +4 3x + 9x = 0 1. x + 3x +3 +(4+ 3)x +(4 3 ) +1=0 13. Find an equation of the parabola with axis = x passing through the points (1, 0), (0, 1), and (1, 1) (a) in ˆx-coordinates, and (b) in x-coordinates.

(, ), θ=90 In Exercises 1, (a) determine whether the conic section is an ellipse, hperbola, or parabola, and (b) perform a rotation, and if necessar a translation, and sketch

Ax 2 Cy 2 Dx Ey F 0. Here we show that the general second-degree equation. Ax 2 Bxy Cy 2 Dx Ey F 0. y X sin Y cos P(X, Y) X

Ax 2 Cy 2 Dx Ey F 0. Here we show that the general second-degree equation. Ax 2 Bxy Cy 2 Dx Ey F 0. y X sin Y cos P(X, Y) X Rotation of Aes ROTATION OF AES Rotation of Aes For a discussion of conic sections, see Calculus, Fourth Edition, Section 11.6 Calculus, Earl Transcendentals, Fourth Edition, Section 1.6 In precalculus