Thomas Calculus 13th [Solutions]

Chapter 11 Practice Exercises 867
57.
2
2 y
y 3x x 4 p 3 p 3 ;
3 4
therefore Focus is 3
4 , 0 , Directrix is x 3
4
58.
2
2 8 y
y x x 4 p 8 p 2 ;
3 3 3
8
3
therefore Focus is 2
3 , 0 , Directrix is x 2
3
59.
2 2
2 y
16x
7y
112 x 1
7 16
2
c 16 7 9 c 3; e c 3
a 4
2
60.
2 2
2 y 2
x 2y 4 x 1 c
4 2
4 2 2
2
c 2; e c
a 2
2
61.
2 2 2 y 2
3x y 3 x 1 c 1 3 4
3
c 2; e c 2
a 1
2;
the asymptotes are y 3x
2
62.
2 2
2 2 2
5y 4x 20 y x 1 c 4 5 9
4 5
c 3, e c 3
a 2
; the asymptotes are y 2 x
5
63.
2
2
x 12y x y 4 p 12 p 3 focus is (0, 3), directrix is y 3, vertex is (0, 0); therefore
12
2
new vertex is (2, 3), new focus is (2, 0), new directrix is y 6, and the new equation is ( x 2) 12( y 3)
64.
2
2 y
y 10x x 4 p 10 p 5 focus is 5
10 2
2 , 0 , directrix is x 5
2 , vertex is (0, 0); therefore
new vertex is 1 , 1 , new focus is (2, 1), new directrix is x 3, and the new equation is
2
2
( y 1) 10 x 1
2
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