{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 3 110 - 260 25 26 27 3 CHAPTER3 PROBLEMS PLUS y =...

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

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 260 25. 26. 27. 3 CHAPTER3 PROBLEMS PLUS y = x4 — 2:132 — a: => 3/ = 4x3 — 4m — 1. The equation of the tangent line at a: = a is y — (a4 — 2(12 — a) — (4a3 4a 1)(ar a) ory — (4(13 — 4a —1)av + (—3114 + 2a2) and similarly for as = b. So if at m : a and m : b we have the same tangent line. then 4a3 # 4a # 1 : 4b3 — 4b — 1 and —3a4 + 2a2 2 —3b4 + 2b2. The first equation gives a3 — b3 = a — b => ((1 — b)(a2 + ab + b2) : (a — b). Assuming a 3A b. we have 1 2 a2 + ab + b2. The second equation gives 3(a4 # b4) : 2(a2 — b2) => 3(a2 — b2)(a2 + b2) : 2(a2 — b2) which is true if a z —b. Substituting into 1 : a2 + ab + b2 gives 1: a2 A a2 + a2 => a : :l:1 so that a = 1 and b = —1 or vice versa. Thus. the points (17 —2) and (—10) have a common tangent line. As long as there are only two such points. we are done. So we show that these are in fact the only two such points. Suppose that a2 — b2 7é 0‘ Then 3(a2 — b2)(a2 + b2) : 2(a2 — b2) gives 3(a2 + b2) : 2 or a2 + b2 : %. 1 Thus,ab: (a2+ab+b2)—(a2+b2) :1—%: %,sob: $.Hence.a2+§% : §.so9a4—t—1:6a2 => 1 0 : 9a4 — 6a2 + 1 : (3a2 7 1)2. So 3(12 1 0 a2 g s b2 9112 % a2, contradicting our assumption that a2 75 b2. Suppose that the normal lines at the three points (a1, (1%). (a2, a2). and (a3, a3) intersect at a common point. Now if one of the ai is 0 (suppose a1 : 0) then by symmetry (12 : —a3, so all + a2 + a3 = 0. So we can assume that none of the ai is 0. The slope of the tangent llne at (0411,01?) IS 2m. so the slope of the normal line 1S ’2— and its equation is (12‘ 1 . , . . y — a? = #_2_ (a: — a7). We solve for the m—coordmate of the 1ntersection of the normal lines from ((11, a?) (1i 1 1 d . 2 : = 2 — -— — = ,2 7 — f :> an (@2112) y a1 2&1 (:0 a1) a2 2112 (er a2) 1 1 2 2 a1 — (12 x —- 7 — : a2 — (11 => :0 2 (—a1 — a2)(a1 + a2) <1} 3: : —2a1a2(a1 + a2) (7%). 2(12 2a1 2a1a2 Similarly, solving for the m-coordinate of the intersections of the normal lines from (a1, a?) and (L137 a3) gives a: : #2a1a3(ai + (13) (T). Equating (7k) and (T) gives a2(a1 + a2) = a3(a1 + a3) 4:» a1(a2 — a3) : (13, — a3 : —(a2 + a3)(a2 — a3) a) a1:#(a2+a3) 41> ai+a2+a3:0. Because of the periodic nature of the lattice points. it suffices to consider the points in the 5 X 2 grid shown. We can see that the minimum value of r occurs when there is a line with slope % which touches the circle centered at (3, 1) ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern