Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

x2 y2 y2 a:2 _ a2 b
61. ;_2 - - = 1 =? - = --- =? y = ±- Jx2- a2.
a b 2 b 2 a 2 a
SECTION 10.5 CONIC SECTIONS 0 31
A= 2 r !!_ J x 2 ·_ a 2 dx ~ 2 b [~ J x2 - a2 - 42 In Jx + J x2 - a2J ] c
la a a 2 2 a·
b
= -[cvc 2 - a 2 -a 2 ln jc + vc 2 - a 2 j + a 2 ln!al)
a
Since a 2 + b 2 = c 2 , c 2 - a 2 = b 2 , and J c2 - a 2 = b.
=!!. [cb- a 2 1n(c + b)+ a 2 !na) = !!_ [ch + a 2 (!na - !n(b +c)))
a
a
= b 2 c/a + abln[a/(b +c)), where c 2 = a 2 + b 2 .
x2
Y2
63. 9x 2 + 4y 2 = 36 '4 + g = 1 =? a= 3, b = 2. By symmetry, x = 0. By Example 2 in Section 7.3, the area o_fthe
top half of the ellipse is H 1rab) = 37T. Solve 9x 2 + 4y 2 = 36 for y to get an equation for the top half of the ell ipse:
9x 2 +4y 2 =36 4y 2 =36-9x 2 ~ y 2 = ~(4-x 2 ) =? y = ~v'4-x 2 .Now
1 1b 1 1 12
1 ( 3 ) 2 3 12
y = - - [f(x)) 2 dx=- - - V 4- x 2 dx= - (4 -x 2 )dx
A a 2 37T - 2 2 2 87T - 2
2 3 [ 1 3] 2 = - · 2 ( 4 - x ) dx = - 4x - - x = -3 ( - 16) = -4
3 ' 12
87T 0 47T 3 0 47T 3 7T
y
so the centroid is (0, 4/ rr).
'ffi . . . I' . I •x 2 y 2 1
65. Dt erenttatmg tmp tctt y, a 2
+ b 2
= =? [y =f. 0]. Thus, the slope ofthe tangent
line at Pis - b:x 1 . The slope of F1P is ___)f];_ and of F2P is ____lt!__ By the formula from Problems Plus, we have
a Y1 X1 + c X1 - c
and
Yl
b 2 Xt
-Xt_+_c + -a2_y_1 _ a 2 yt + b 2 x 1(x1 +c) _ a 2 b 2 + b 2 cx1
tan a = _.:........;.---:-:.-----'-'-=--
1 _ b 2 x1Y1 - a 2 y1 (x1 + c) - b 2 X tY1 - C 2 XtY1 + a 2 CIJ1
a 2 yt (Xt +c)
b 2 (cxt+a 2 ) · b 2
= CIJt(CXt + a 2 ) = cy1
- a 2 y'f. - b 2 x1 (x1 -c) - a 2 b 2 + b 2 cx1 b 2 (ext - a 2 ) b 2
= a 2 y1 (x1- c) - b 2 X 1Y1 = c 2 x1y1 - a 2 C1Jt = CIJ1 (cx1 - a 2 ) = CIJ1
Thus, a = {3.
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