# Howard, Anton - Cálculo - 6ª Edição - Exercicios Resolvidos

(Parte 3 de 6)

(b) x y

4. (a) y F(0, 1)

5. (a)

V(2, 3)

(b) -4 x y

 94F(–2, – )

(b)

 52V(2, )

F(2, 2) y = 3

(b)

(b)

 1716F(–1, )

xy V (–1, 1)

(4, 0) (0, 3)

(0, 3)

(0, 5)

(0, 2)

(3, 0) x

Exercise Set 1.4 463

(1, 6)

(1, 0) x

(5, 3)(–3, 3)

(1, 5) (–3, 7)

(0, 1)(0, –2 + √5) x y

(2, 1) x

(1, 3) (–1, 6) x y

 y = x

(0, 2) x

 y = x

(0, 3) x

 35y = – x35y = x
 45y = – x45y = x

(5, 4) x

 y – 4 = (x – 2)2

x y

 y + 4 = (x – 2)
 y + 4 = – (x – 2)

x y

Exercise Set 1.4 465

(1, 1) x

 y – 1 = (x + 1)1

(-2, 5)

(-2, 1) x

 y – 3 = – (x + 2)23 y – 3 = (x + 2)
 y + 5 = – (x + 2)
 y + 5 = (x + 2)

(b) The vertex is half way between the focus and directrix so the vertex is at (2,4), the focus is 3 units to the left of the vertex so p =3 , (y − 4)2 = −12(x − 2)

(10, 12) x

Exercise Set 1.4 467

(b) As in Part (a), y = −4h

3 bh

41. We may assume that the vertex is (0,0) and the parabola opens to the right. Let P(x0,y0)b ea point on the parabola y2 =4 px, then by the deﬁnition of a parabola, PF = distance from P to

42. Let p = distance (in millions of miles) between the vertex (closest point) and the focus F, then PD = PF ,2 p + 20 = 40, p = 10 million miles.

Directrix

43. Use an xy-coordinate system so that y2 =4 px is an equation of the parabola, then (1,1/2) is a point on the curve so (1/2)2 =4 p(1), p =1 /16. The light source should be placed at the focus which is 1/16 ft. from the vertex.

(2, 1) x

Use either equation to ﬁnd that y = ±√ 3i f x =2

or if x = −2. The curves intersect at

45. (a) P :( bcost,bsint); Q :( acost,asint); R :( acost,bsint)

(b) For a circle, t measures the angle between the positive x-axis and the line segment joining the origin to the point. For an ellipse, t measures the angle between the x-axis and OPQ, not OR.

46. (a) Fora ny point( x,y),the equation y = bsinht has a unique solution t, −∞ <t< +∞. On the hyperbola,

(b) 3

47. (a) Fora ny point( x,y), the equation y = btant has a unique solution t where −π/2 <t<π /2.

On the hyperbola, x2

(Parte 3 de 6)