Pre-Calc Homework Solutions 464

Pre-Calc Homework Solutions 464 - 464 Appendix A5.3 10. 3x...

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Unformatted text preview: 464 Appendix A5.3 10. 3x 2 6xy 3y 2 4x 5y 12 B 2 4AC 62 4(3)(3) 0 Parabola 11. 3x 2 5xy 2y 2 7x B 2 4AC ( 5)2 12. 2x 2 4.9xy 3y 2 4x B 2 4AC ( 4.9)2 Hyperbola c 27. continued From the preliminary result in Exercise 26, x0b 2 y0a 2 x0 x0 y0 c y0 14y 1 4(3)(2) 1 7 4(2)(3) 0 Hyperbola 0.01 0 tan 1 x02b 2 x0y0a 2 a 2b 2 x0y0c 2 x0b 2 y0a 2 x0 b 2 c y0 a 2 c x0b 2c y0a 2c y0 x0 c y0 x0 c y02a 2 x0y0b 2 b2 . In a similar manner, y0c x0b 2 y0a 2 x0b 2 13. x 2 3xy 3y 2 6y 7 B 2 4AC ( 3)2 4(1)(3) 21xy 14. 25x B 2 4AC 2 2 3 41 0 Ellipse 0 Hyperbola 0 Ellipse 4y 212 2 2 350x 0 4(25)(4) 3xy 2y 17y 2 0 15. 6x B 2 4AC 32 4(6)(2) 39 16. 3x 2 12xy 12y 2 B 2 4AC 122 17. cot 2 tan , and therefore x x 2 2 1 2 2 2 2 A B C 0 1 435x 9y 72 0 4(3)(12) 0 Parabola 02 2 tan 1 y0a 2 b2 . Since tan y0c x sin 2 2 4 ; y cos x cos x 2 2 1 2 y 2 2 2 y sin , y y ,y 2 2 2 2 and are acute angles, we have . APC, and the APB. Since APC x 2 y 28. The tangent to the ellipse of P bisects tangent to the hyperbola at P bisects and APB are a linear pair, so that m APB m LAPB 2 x y x 2 2 y 2 4 Hyperbola ; x 18. cot 2 2x 1 1 1 y 2 A B C 02 y sin, y m APC m APC 2 180 and therefore x 90 , the tangents to the ellipse and x 2 2 2 2 2 4 x cos x 2 2 2 2 2 x sin 2 2 2 2 y cos hyperbola are perpendicular. y ,y 2 2 2 2 2 2 x y y 2 2 2 x 2 y 2 x 1 x 2 2 y s Appendix A5.3 (pp. 612618) 1. x 3xy y B 2 4AC 2 2 2 x y x x 0 ( 3)2 2 4(1)(1) 5 0 Hyperbola 1 2 1 2 1 2 1 2 1 2 xy y x y x xy 2 2 2 2 2 1 2 3 1 2 y 1 x2 y 1 3x 2 y 2 2 2 2 2 18xy 27y 5x 7y 4 2. 3x B 2 4AC ( 18)2 4(3)(27) 0 Parabola 3. 3x 2 7xy B 2 4AC Ellipse 4. 2x 2 B2 15xy 4AC 17y 2 1 ( 7)2 4(3) 17 2y 2 x y 0 ( 15)2 4(2)(2) 0.477 0 Ellipse A B C 3 1 1 3 19. cot 2 2 2 3 3 6 ; therefore x 1 0 Ellipse x 3 2 3 2 3 x cos x 1 y ,y 2 1 y 2 2 y sin , y 1 x 2 3 2 x sin y y cos 5. x 2 2xy y 2 2x y 2 B 2 4AC 22 4(1)(1) 0 0 Parabola x 3 2 6. 2x 2 y 2 4xy 2x 3y 6 B 2 4AC 42 4(2)( 1) 24 0 Hyperbola 7. x 2 4xy 4y 2 3x 6 B 2 4AC 42 4(1)(4) 8. x 2 y 2 3x B 2 4AC 9. xy y 2 3x Hyperbola 2y 02 10 4(1)(1) 4AC 0 Parabola 4 12 0 Ellipse (circle) 4(0)(1) 1 0 4x 2 1 x 2 3 x 3 1 y 2 2 1 x 2 3 2 3 y 2 y 3 2 8 y 2 x 1 y 2 8 2 3 x 16y 1 2 0 5 B2 0 Parabola ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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