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
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 A C @ @ = 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).
ROTATION OF AES 3 EAMPLE 3 Identif and sketch the curve 73 7 3 7 SOLUTION This equation is in the form of Equation with A 73, 7, and C. Thus cot A C 73 7 7 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 7 7 3 3 3 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 73 +7+ +3--7= or @+(-)@=
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.. 6. 7. 8. s3 9. 97 9 3. 3 s 6 9. s3 s3. 6 8s (8s 3) (6s ) 7 3. (a) Use rotation of aes to show that the equation 36 96 6 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 7 3 8 6 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.
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) 6 8 7 (, 7 6), ( 7, 8)
6 ROTATION OF AES Solutions: Rotation of Aes. = cos 3 +sin3 =+ 3, = sin 3 +cos3 = 3. 3. = 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 + = /3 + =. [An ellipse, center (, ),focion -ais with a =, b = 6/3, c = 3/3.] 9. cot θ = 97 3 9 = 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 ) + 9 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 ].
ROTATION OF AES 7 3. (a) cot θ = A C = 7 3 +3 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 = 3 7 6 = 7 and = + 7 3 6 =. 8 (c) The directri is = 6,so + 3 = 6 6 8 +7=.. 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).