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1 ROTATION OF AES For a discussion of conic sections, see Review of Conic Sections 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 B C D E F can be analzed b rotating the aes so as to eliminate the term B 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 FIGURE P(, ) P(, ) r cos r cos r sin 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 B solving Equations for and we obtain cos sin sin cos r sin r P EAMPLE If the aes are rotated through 6, find the -coordinates of the point whose -coordinates are, 6 6 SOLUTION Using Equations with, 6, and, we have cos 6 6 sin 6 s Thomson Brooks-Cole copright 7 sin 6 6 cos 6 s The -coordinates are ( s, s)

2 ROTATION = or - = Now let s tr to determine an angle such that the term B in Equation disappears when the aes are rotated through the angle If we substitute from Equations in Equation, we get E sin cos F Epanding and collecting terms, we obtain an equation of the form where the coefficient B of is To eliminate the term we choose so that B, that is, or EAMPLE Show that the graph of the equation is a hperbola SOLUTION Notice that the equation is in the form of Equation where A, B, and C According to Equation, the term will be eliminated if we choose so that A cos sin B cos sin sin cos C sin cos D cos sin A B C D E F B C A sin cos Bcos sin C A sin B cos A C sin cos cot A C B cot A C B 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 Thomson Brooks-Cole copright 7 FIGURE 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 EAMPLE Identif and sketch the curve SOLUTION This equation is in the form of Equation with A 7, B 7, and C Thus cot A C B FIGURE From the triangle in Figure we see that The values of cos and sin can then be computed from the half-angle formulas: The rotation equations () become Substituting into the given equation, we have which simplifies to Completing the square gives and we recognize this as being an ellipse whose center is, in -coordinates Since, we can sketch the graph in Figure cos ( ) 7 7( ) 7( )( ) ( ) or cos sin ( ) ( ) 7 cos 7 cos cos 7 7 (, ) Å7 FIGURE = Thomson Brooks-Cole copright 7

4 ROTATION OF AES EERCISES A Click here for answers S Click here for solutions (c) Find an equation of the directri in the -coordinate sstem Find the -coordinates of the given point if the aes are rotated through the specified angle,,,,,, 6,, Use rotation of aes to identif and sketch the curve s s 6 9 s s 6 8s (8s ) (6s ) 7 (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 (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 B AC B AC 7 Use Eercise 6 to show that Equation represents (a) a parabola if B AC, (b) an ellipse if B AC, and (c) a hperbola if B 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 Thomson Brooks-Cole copright 7

5 ROTATION OF AES ANSWERS ((s ), (s )) s, parabola 7, ellipse S Click here for solutions (s, s ) 9 9, ellipse, hperbola (a) (b) (c) sin! (, 6), 7 ( 7, 8) Thomson Brooks-Cole copright 7

6 6 ROTATION OF AES SOLUTIONS = cos +sin =+, = sin +cos = =cos +sin = 7, = sin +cos = = cos6 +sin6 = +, =sin6 +cos6 = + Using the half-angle formulas [Equation 7 in Appendi A], we have cos = +cos = cos +sin = = +( /) = + cos = + and similarl sin = Thus, + +, = sin +cos = 6 + [Another method: Using the identit cos θ +sinθ = cos(θ ), we get the simpler epression = cos +sin = cos( )= = 6 Similarl, =cos sin = cos( + )= cos6 = ] cot θ = A C B = θ = π θ = π [b Equations ] = and = + Substituting these into the curve equation gives =( ) ( + ) = or = [Parabola, verte (, ), directri = /, focus /, ] 6 cot θ = A C B = θ = π [Equations ] =, = + Substituting gives =or + =, which is an ellipse centered at (, ), foci on the -ais, / a =, b = 6,andc = 7 cot θ = A C B = θ = π θ = π [b Equations ] curve equation gives = + = and = + Substituting these into the = / + = [An ellipse, center (, ),focion -ais with a =, b = 6/, c = /] Thomson Brooks-Cole copright 7

7 ROTATION OF AES 7 8 cot θ = = θ = π = and = + Substituting and simplifing gives 6 =,or / =, a hperbola with foci on the -ais, a = 6, b =, c = 6, and center (, ) 97 9 cot θ = = 7 9 tan θ = π < θ <π 7 and cos θ = 7 π <θ< π, cos θ =, sin θ = = cos θ sin θ = and + = sin θ + cos θ = Substituting, we get 97 ( ) + 9 ( )( + )+ ( + ) =, which simplifies to + =(anellipse with foci on -ais, centered 9 at origin, a =, b =) cot θ = A C B = 6 = cos θ = 9 +(/9) cos θ = = and sin θ = = = + Substituting into the given equation and simplifing we get =or =, a hperbola with foci on the -ais, center at (, ), a =, b =,andc = 6 cot θ = A C B = θ = π 6 = + =, and Substituting into the curve equation and simplifing gives 8 = ( ) =[a hperbola with foci on -ais, centered at (, ), a =,b=/, c =/ ] Thomson Brooks-Cole copright 7

8 8 ROTATION OF AES cot θ = 6 8 = 7 sin θ = cos θ = 7 9 cos θ =, = Substituting gives =6 and = =7 = ( +),sothisisa parabola with verte,, directri = 9,andfocus 7, 8 8 (a) cot θ = A C B = 7 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 6 = 7 and = = 8 (c) The directri is = 6,so + = = (a) cot θ = 7 [as in Eample ] = + and = [substituting and simplifing] + = ( +) =a hperbola with center (, )=(, ) andfociontheline = (b) a =, b = c = 6 -coordinates of foci are, ± 6 -coordinates are = ± 6 and = + ± 6 giving 8 6, and 8 + 6, 6 6 (c) -coordinates of vertices are, ± -coordinates are = ± and = + ± giving 8, 6 + and 8 +, 6 (d) The asmptotes are = ± a b ( +)=± + = ± + + (e) e = c 6/ a = / = ( +)so b () the equations in -coordinates are ± + + ± / ± = A rotation through θ changes Equation to A( cos θ sin θ) + B( 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 Thomson Brooks-Cole copright 7

9 ROTATION OF AES 9 6 From the solution to Eercise, we see that (B ) A C =(C A) sin θ +B(C A)sinθcos θ + B cos θ (A cos θ + B sin θ cos θ + C sin θ) (A sin θ B sin θ cos θ + C cos θ) Multipling this out and using the identities cos θ sin θ =cosθ and sinθcos θ =sinθ,itsimplifies to B (cos θ +sin θ) AC( sin θ cos θ) AC cos θ +sin θ = B AC(cos θ +sin θ) = B AC 7 Choose θ so that B =ThenB AC =(B ) A C = A C ButA 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 : B AC =for a parabola, B AC > for a hperbola, B 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) 8 In #9, B AC =, <,soanellipse In #, B AC = 68 >, so a hperbola In #, B AC =>, so a hperbola In #, B AC =,soaparabola Thomson Brooks-Cole copright 7