Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.5 CONIC SECTIONS ¤ 441
(x 0 − a) 2 = a 2 +4p 2 ⇔ x 0 = a ± a 2 +4p 2 . The slopes of the tangent lines at x = a ± a 2 +4p 2
are a ± a 2 +4p 2
, so the product of the two slopes is
2p
a + a 2 +4p 2
2p
· a − a 2 +4p 2
2p
= a2 − (a 2 +4p 2 )
4p 2
= −4p2
4p 2 = −1,
showing that the tangent lines are perpendicular.
59. For x 2 +4y 2 =4,orx 2 /4+y 2 =1, use the parametrization x =2cost, y =sint, 0 ≤ t ≤ 2π to get
L =4 π/2
0
(dx/dt)2 +(dy/dt) 2 dt =4 π/2
0
4sin 2 t +cos 2 tdt=4
π/2
0 3sin 2 t +1dt
Using Simpson’s Rule with n =10, ∆t = π/2 − 0
10
= π 20 ,andf(t) = 3sin 2 t +1,weget
L ≈ 4 3
π
20
f(0) + 4f
π
20
+2f
2π
20 + ···+2f 8π
20 +4f 9π
20 + f π
2 ≈ 9.69
61.
x 2
a − y2
y2
=1 ⇒ 2 b2 b = x2 − a 2
⇒ y = ± b √
x2 − a 2 a 2 a
2 .
A =2
c
a
b
a
x2 − a 2 dx 39 = 2b
a
x
2
x2 − a 2 − a2
2 ln x +
x2 − a 2
c
a
= b √
c c2 − a
a
2 − a 2 ln √ c + c2 − a 2 + a 2 ln |a|
Since a 2 + b 2 = c 2 ,c 2 − a 2 = b 2 ,and √ c 2 − a 2 = b.
= b cb − a 2 ln(c + b)+a 2 ln a = b cb + a 2 (ln a − ln(b + c))
a
a
= b 2 c/a + ab ln[a/(b + c)], wherec 2 = a 2 + b 2 .
63. Differentiating implicitly, x2
a + y2
2x
=1 ⇒ 2 b2 a + 2yy0 =0 ⇒ y 0 = − b2 x
2 b 2 a 2 y
[y 6= 0]. Thus, the slope of the tangent
line at P is − b2 x 1
a 2 y 1
. The slope of F 1 P is
we have
tan α =
1 −
and tan β =
1 −
Thus, α = β.
y 1
x 1 + c + b2 x 1
a 2 y 1
b 2 x 1 y 1
a 2 y 1 (x 1 + c)
= b2 cx 1 + a 2
cy 1 (cx 1 + a 2 ) = b2
cy 1
− b2 x 1
a 2 y 1
− y 1
x 1 − c
b 2 x 1 y 1
a 2 y 1 (x 1 − c)
y 1
x 1 + c and of F 2P is
y 1
. By the formula in Problem 17 on text page 268,
x 1 − c
= a2 y1 2 + b 2 x 1 (x 1 + c)
= a2 b 2 + b 2 cx 1 using b 2 x 2 1 + a2 y1 2 = a2 b 2 ,
a 2 y 1 (x 1 + c) − b 2 x 1 y 1 c 2 x 1 y 1 + a 2 cy 1 and a 2 − b 2 = c 2
= −a2 y 2 1 − b 2 x 1 (x 1 − c)
a 2 y 1 (x 1 − c) − b 2 x 1 y 1
= −a2 b 2 + b 2 cx 1
c 2 x 1 y 1 − a 2 cy 1
= b2 cx 1 − a 2
cy 1 (cx 1 − a 2 ) = b2
cy 1