Pre-Calc Homework Solutions 462

Pre-Calc Homework Solutions 462 - 462 Appendix A5.2 4)2 y2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 462 Appendix A5.2 4)2 y2 2 x 3 1 c 2 14. Using PF (x x2 x2 15. 9x 2 c 5 9 e PD, we have 2 (x 9 22. Focus ( 2, 0) and directrix x a e 1 ae 2 2 e 1 2 2 2 e 1 e2 2 ae 2. 2x 2 and 4) y 2 4 (x 9 9) 2 4e (y 0)2 8x y2 16 y2 4 2 (x 9 18x 81) x2 36 y2 20 Then PF 1. (x x2 c a 5 ; 4 2PD y2 4 y2 (y 1 (x 4 x 2)2 1 2 2 1 2 20 5x 2 144 x2 16 9y 2 y2 9 180 or 1 5e 2)2 4x 16y 2 a 2 y2 4 x2 y 3 x 2 1 4 b 2 16 9 23. (x 3x 2 1)2 2x 3 x2 3)2 y2 8x 1) (x 1) 45 3 y 2 1 2 asymptotes are y x 16. y 2 c 0 x2 a e 16 . 5 3 x; F( 5, 0); directrices are 4 x2 x2 8 6y 60y 5(y 2 9 4 12y 9 2 (y 4 4y 4) 8 a2 y2 8 1 8 8 4e c a 4 8 4x 2 2; 4(x 2 24. c 2 1 c2 thus, 10 2x; 2 5 2 10 x a2 2 5y 2 2x 2 0 36) 4 4 180 b2 (y asymptotes are y y 17. 8x 2 c e F( 18. 8y 2 c e F(0, c a c a x; F(0, 8 2 x2 2 4); directrices are 6) 36 2 1 c2 e2a 2 x a2 2 0 a e 2 y2 8 a2 b2 b2 e2a 2 b2 y b2 2 a 2; e a2 y a 2(e2 2 c c a ea 1); 2y 2 a2 16 b2 10 2 a 2(e2 1) 2 8 1 1; the asymptotes of this hyperbola are y 5; asymptotes are y 0 1 8 10 x ; 2 2 5 c a a e e2 1x. As e increases, the slopes of the asymptotes increase and the hyperbola approaches a single straight line. 25. The ellipse must pass through (0, 0) c ( 1, 2) lies on the ellipse a 2b 0; the point 8. The ellipse is 0 10, 0); directrices are x 2x 2 a2 16 b2 10 2 y 2 2 x 8 2 2 5; asymptotes are y 0 a e tangent to the x-axis its center is on the y-axis, so a 2 10 10); directrices are y 1) and e 3 b2 1 3c a2 9 3a c2 a2 and b 4x 2 x2 y2 (y 4 the equation is 4x 2 4y 2) 4 2 y2 2)2 4y 4 0. Next, 19. Vertices (0, c y2 3a x2 8 1 and e 9 1 8 3 4 4 4x 2 1a (y 2 and b a2 3 1 (now using the 4 1 3 standard symbols) c2 c 3 and e 1 c a b2 3 2 3e 20. Foci ( 3, 0) and e a 1 b2 c2 3c y 8 2 c a 2 . 3a 1 8 x2 ae 2. (y 0)2 21. Focus (4, 0) and directrix x ae e2 2c 2e 4)2 2 4 and a e 2 2 4 e2 2 e2 (x 2) Then PF (x x2 x2 4) 2 2PD y 2 2x 2 2(x y2 8 x 8 2 8x y2 16 2(x 2 y 8 2 4x 1 4) ...
View Full Document

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.

Ask a homework question - tutors are online