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 P(X, Y) X

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1 Rotation of Aes For a discussion of conic sections, see Appendi. In precalculus or calculus ou ma have studied conic sections with equations of the form A C D E F Here we show that the general second-degree equation A C D E F can be analzed b rotating the aes so as to eliminate the term. In Figure the and aes have been rotated about the origin through an acute angle to produce the and aes. Thus, a given point P has coordinates, in the first coordinate sstem and, in the new coordinate sstem. To see how and are related to and we observe from Figure that r cos r cos r sin r sin P(, ) P(, ) r P FIGURE FIGURE The addition formula for the cosine function then gives r cos rcos cos sin sin r cos cos r sin sin cos sin A similar computation gives in terms of and and so we have the following formulas: cos sin sin cos solving Equations for and we obtain 3 cos sin sin cos EAMPLE If the aes are rotated through 6, find the -coordinates of the point whose -coordinates are, 6. SOLUTION Using Equations 3 with, 6, and, we have cos 6 6 sin 6 3s3 sin 6 6 cos 6 s3 3 The -coordinates are ( 3s3, 3 s3). 6

2 ROTATION OF AES Now let s tr to determine an angle such that the term in Equation disappears when the aes are rotated through the angle. If we substitute from Equations in Equation, we get A cos sin cos sin sin cos C sin cos D cos sin E sin cos F Epanding and collecting terms, we obtain an equation of the form A C D E F where the coefficient of is C A sin cos cos sin C A sin cos To eliminate the term we choose so that, that is, A C sin cos or cot A C EAMPLE Show that the graph of the equation is a hperbola. SOLUTION Notice that the equation is in the form of Equation where A,, and C. According to Equation, the term will be eliminated if we choose so that cot = or - = This will be true if, that is,. Then cos sin s and Equations become s s s s π Substituting these epressions into the original equation gives s s or s s FIGURE 3 We recognize this as a hperbola with vertices (s, ) in the -coordinate sstem. The asmptotes are in the -sstem, which correspond to the coordinate aes in the -sstem (see Figure 3).

3 ROTATION OF AES 3 EAMPLE 3 Identif and sketch the curve SOLUTION This equation is in the form of Equation with A 73, 7, and C. Thus cot A C From the triangle in Figure we see that cos 7 7 FIGURE The values of cos and sin can then be computed from the half-angle formulas: cos sin The rotation equations () become cos cos Substituting into the given equation, we have 73( 3 ) 7( 3 )( 3 ) ( 3 ) 3( 3 ) ( 3 ) 7 which simplifies to 3 Completing the square gives or and we recognize this as being an ellipse whose center is, in -coordinates. Since, we can sketch the graph in Figure. cos ( ) 37 (, ) Å37 FIGURE =

4 ROTATION OF AES Eercises A Click here for answers. S Click here for solutions. Find the -coordinates of the given point if the aes are rotated through the specified angle..,, 3., 3, 3.,, 6.,, Use rotation of aes to identif and sketch the curve s s 6 9. s3 s3. 6 8s (8s 3) (6s ) 7 3. (a) Use rotation of aes to show that the equation represents a parabola. (b) Find the -coordinates of the focus. Then find the -coordinates of the focus. (c) Find an equation of the directri in the -coordinate sstem.. (a) Use rotation of aes to show that the equation represents a hperbola. (b) Find the -coordinates of the foci. Then find the -coordinates of the foci. (c) Find the -coordinates of the vertices. (d) Find the equations of the asmptotes in the -coordinate sstem. (e) Find the eccentricit of the hperbola.. Suppose that a rotation changes Equation into Equation. Show that A C A C 6. Suppose that a rotation changes Equation into Equation. Show that AC AC 7. Use Eercise 6 to show that Equation represents (a) a parabola if AC, (b) an ellipse if AC, and (c) a hperbola if AC, ecept in degenerate cases when it reduces to a point, a line, a pair of lines, or no graph at all. 8. Use Eercise 7 to determine the tpe of curve in Eercises 9.

5 ROTATION OF AES S Answers Click here for solutions.. ((s3 ), (s3 )) 3. (s3, s3 ). s, parabola 9. 9, ellipse sin!. 3, hperbola 7. 3, ellipse 3. (a) (b) (c) (, 7 6), ( 7, 8)

6 6 ROTATION OF AES Solutions: Rotation of Aes. = cos 3 +sin3 =+ 3, = sin 3 +cos3 = = cos 6 +sin6 = + 3, =sin6 +cos6 = 3+.. cot θ = A C = θ = π θ = π [b Equations ] = and = +. Substituting these into the curve equation gives =( ) ( + ) = or =. [Parabola, verte (, ), directri = /, focus /, ]. 7. cot θ = A C = θ = π θ = π [b Equations ] = curve equation gives = + and = + +. Substituting these into the = /3 + =. [An ellipse, center (, ),focion -ais with a =, b = 6/3, c = 3/3.] 9. cot θ = = 7 tan θ = 7 π < θ <π and cos θ = 7 π <θ< π, cos θ = 3, sin θ = 3 = cos θ sin θ = and +3 = sin θ + cos θ =. Substituting, we get 97 (3 ) (3 )( +3 )+ ( +3 ) =, which simplifies to + =(anellipse with foci on -ais, centered 9 at origin, a =3, b =).. cot θ = A C = 3 θ = π 6 3 =, = + 3. Substituting into the curve equation and simplifing gives 8 = ( ) 3 =[a h p e r b ol a w it h f oc i on -ais, centered at (, ), a =,b=/ 3, c =/ 3 ].

7 ROTATION OF AES 7 3. (a) cot θ = A C = so, as in Eercise 9, = and =. Substituting and simplifing we get += =, which is a parabola. (b) The verte is (, ) and p =,sothe -coordinates of the focus are, 7 6 6,andthe-coordinates are = = 7 and = =. 8 (c) The directri is = 6,so + 3 = =.. A rotation through θ changes Equation to A( cos θ sin θ) + ( cos θ sin θ)( sin θ + cos θ)+c( sin θ + cos θ) + D( cos θ sin θ) +E( sin θ + cos θ)+f =. Comparing this to Equation, we see that A + C = A(cos θ +sin θ)+c(sin θ +cos θ)=a + C. 7. Choose θ so that =.Then AC =( ) A C = A C.utA C will be for a parabola, negative for a hperbola (where the and coefficients are of opposite sign), and positive for an ellipse (same sign for and coefficients). So : AC =for a parabola, AC > for a hperbola, AC < for an ellipse. Note that the transformed equation takes the form A + C + D + E + F =, or b completing the square (assuming A C 6=), A ( ) + C ( ) = F,sothatifF =, the graph is either a pair of intersecting lines or a point, depending on the signs of A and C.IfF 6=and A C >, then the graph is either an ellipse, a point, or nothing, and if A C <, the graph is a hperbola. If A or C is, we cannot complete the square, so we get A ( ) + E + F =or C ( ) + D + F =. This is a parabola, a straight line (if onl the second-degree coefficient is nonzero), a pair of parallel lines (if the first-degree coefficient is zero and the other two have opposite signs), or an empt graph (if the first-degree coefficient is zero and the other two have the same sign).