Thomas Calculus 13th [Solutions]

Section 11.6 Conic Sections 845
56.
2
y
3
2
x 1 center is (0, 0), vertices are 0, 3 and 0, 3 ;
(0, 2) and (0, 2); the asymptotes are
2 2
c a b 3 1 2 foci are
y
x y 3 x ; therefore the new center is (1, 3), the new
3
vertices are 1, 3 3 , and the new foci are (1, 5) and (1,1); the new asymptotes are y 3 3( x 1); the
new equation is
2
( y 3) 2
( x 1) 1
3
57.
2 2 2 2 2 2
x 4x y 12 x 4x 4 y 12 4 ( x 2) y 16; this is a circle: center at C( 2, 0), a 4
58.
2 2 2 2 2 2
2x 2y 28x 12y 114 0 x 14x 49 y 6y 9 57 49 9 ( x 7) ( y 3) 1; this is
a circle: center at C(7, 3), a 1
59.
2 2 2
x 2x 4y 3 0 x 2x 1 4y 3 1 ( x 1) 4( y 1); this is a parabola: V ( 1,1), F( 1, 0)
60.
2 2 2
y 4y 8x 12 0 y 4y 4 8x 12 4 ( y 2) 8( x 2); this is a parabola: V ( 2, 2), F(0, 2)
61.
2 2 2 2 2 2 ( x 2) 2
x 5y 4x 1 x 4x 4 5y 5 ( x 2) 5y 5 y 1; this is an ellipse: the center
5
2 2
is ( 2,0), the vertices are 2 5, 0 ; c a b 5 1 2 the foci are ( 4, 0) and (0, 0)
2
62.
2 2 2 2 2 2
2 ( y 3)
9x 6y 36 y 0 9x 6 y 6y 9 54 9x 6( y 3) 54 x
2
1; this is an ellipse:
6 9
the center is (0, 3) the vertices are (0, 0) and (0, 6);
0, 3 3
2 2
c a b 9 6 3 the foci are
63.
2 2 2 2 2 2 ( x 1) 2
x 2y 2x 4y 1 x 2x 1 2 y 2y 1 2 ( x 1) 2( y 1) 2 ( y 1) 1;
2
this is an ellipse: the center is (1, 1), the vertices are 1 2, 1 ;
(2, 1) and (0, 1)
2 2
c a b 2 1 1 the foci are
2
64.
2 2 2 2 2 2 2 ( y 1)
4x y 8x 2y 1 4 x 2x 1 y 2y 1 4 4( x 1) ( y 1) 4 ( x 1) 1;
4
this is an ellipse: the center is ( 1, 1) the vertices are ( 1, 3) and ( 1, 1);
foci are 1, 1 3
2 2
c a b 4 1 3 the
2
65.
2 2 2 2 2 2
x y 2x 4y 4 x 2x 1 y 4y 4 1 ( x 1) ( y 2) 1; this is a hyperbola: the center
is (1, 2), the vertices are (2, 2) and (0, 2);
asymptotes are y 2 ( x 1)
2 2
c a b 1 1 2 the foci are 1 2, 2 ; the
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