Business Calc Homework w answers_Part_93

And are both less than 90 we have m1 1 m2 m1m2 to

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: an 1 tan 1 2 2 m1 1 m2 m1 m2 tan 2 tan 27. To prove the reflective property for hyperbolas: b 2x 2 2b 2x 2 and m2 tan .) a 2y 2 2a 2yy b x a 2y a 2b 2 0 Now we prove the reflective property of ellipses (see the accompanying figure): 2b x 2 y b 2x . a2y Let P(x0, y0) be a point of tangency (see the accompanying figure). The slope from P to F( c, 0) is and from P to F2(c, 0) it is y0 x0 c y0 x0 c 2a yy 2 0 y Let P(x0, y0) be any point on the ellipse y (x0) bx0 a a 2 . Let the tangent through b x0 x02 a 2y0 1 2 P meet the x-axis in point A, . Let F1(c, 0) and y0 x0 c and define the angles will show that tan y F1PA tan . and F2 PA . We F2( c, 0) be the foci. Then mPF mPF y0 x0 c 2 and . Let and be the angles between the tangent line and PF1 and PF2 respectively. Then b 2x0 y0 x0 b x0y0 a 2y0(x0 c) 2 tan 1 a 2y0 c F1(c, 0) P(x0, y0) A x F2(c, 0) b 2x02 a y0x0 b 2x0c a y0 c b 2x0c a y0c 2 2 2 b 2x0c a y0c (b 2x02 (a 2 2 a 2y02 b 2x0y0 a 2y02) b 2)x0y0 b2 . cy0 a 2b 2 c x0y0 2 464 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, s...
View Full Document

This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

Ask a homework question - tutors are online