Chapter 13 Review Exercises
80. Rewrite the first equation.
2
log 3
2log 3
xy
xy
=+
=+
Substitute 1 for logyx−in the first equation and
81. 32 6xy−≤
83.
2
32
243
3 2
612 9
xx
x
xx x
−+
−
−+
84. For
1
1
(1) (1) 1 1
1; 31 2
3131
n
n
n−−−
====−
−
−−
For
2
2
(1) (1) 1 1
2; 91 8
3131
n
n
n−−
====
−
−−
For
22
6; 1 6 1 36 1 37nn=+=+=+=
2 5 10 17 26 37 97++ + + + =
3.
()()
22
2
1
001
xy
xy
+=
−+−=
The center is
()
0, 0 and the radius is 1 unit.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
4.
()()
()
()
()
22
222
239
233
xy
xy
++−=
−− + − =
5.
()( )
22
22
4240
4 2 4
xy xy
xx yy
+−+−=
−++ =
Complete the squares.
()
22
2
424
22
b−
==−=
The center is
()
2, 1− and the radius is 3 units.
6.
()
22
22
40
4 0
xy y
xy y
+−=
+− =
Complete the square.
The center is
()
0, 2 and the radius is 2 units.
vertices are
() ()
6,0 and 6,0 .− Since
2
25,b=
5b= and endpoints of the minor axis are
() ()
0, 5 and 0, 5 .−
Chapter 13 Review Exercises
8.
22
1
25 16
xy
+=
Because the denominator of the
2
termx− is greater
than the denominator of the
2
term,y− the major
9.
22
22
22
416
416
16 16 16
1
416
xy
xy
xy
+=
+=
+=
Because the denominator of the
2
termy− is
10.
22
22
22
49 36
49 36
36 36 36
1
94
xy
xy
xy
+=
+=
+=
11.
()()
22
12
1
16 9
xy−+
+=
The center of the ellipse is
()
1, 2 .− Because the
denominator of the
2
termx− is greater than the
denominator of the
2
term,y− the major axis is
Chapter 13 Conic Sections and Systems of Nonlinear Equations
12.
()()
22
12
1
916
xy+−
+=
The center of the ellipse is
()
1, 2 .− Because the
denominator of the
2
termy− is greater than the
Center Vertices
Endpoints of
Minor Axis
()
1, 2−
()
()
1, 2 4
1, 2
−−
=− −
()
()
13,2
4, 2
−−
=−
13. From the figure, we see that the major axis is
horizontal with a = 25, and b = 15.
22
22
1
25 15
xy
+=
()
()
2
2
2
196
5625 5625 1
625 225
9 196 25 5625
1764 25 5625
y
y
y
+=
+=
+=
The height of the archway 14 feet from the center is
approximately 12.43 feet. Since the truck is 12 feet
high, the truck will clear the archway.
2
2
xy
extended diagonals to obtain the asymptotes. Graph
the hyperbola.
Chapter 13 Review Exercises
15.
2
2
22
1
16
1
16 1
yx
yx
−=
−=
16.
22
22
22
916 144
9 16 144
144 144 144
1
16 9
xy
xy
xy
−=
−=
−=
17.
22
22
22
416
416
16 16 16
1
yx
yx
yx
−=
−=
−=
Graph the hyperbola.
18.
()
2
34xy=− −
This is a parabola of the form
()
2
xayk h=−+.
Since 1a= is positive, the parabola opens to the
right. The vertex of the parabola is
()
4, 3 .− The
Chapter 13 Conic Sections and Systems of Nonlinear Equations
The y–intercepts are 1 and 5.
19.
()
2
232xy=− + +
This is a parabola of the form
()
2
xayk h=−+.
Since 2a=− is negative, the parabola opens to the
The x–intercept is −16. Replace x with 0 to find the
y–intercepts.
()
()
2
2
02 3 2
02 692
y
yy
=− + +
=− + + +
The y–intercepts are –4 and –2.
20.
2
812xy y=−+
This is a parabola of the form
2
xay byc=++.
16 32 12 4.
=−+=−
The vertex of the parabola is
()
4, 4 .− The axis of
symmetry is 4.y=
()()
062
yy
=− −
60 or 20
62
yy
yy
−= −=
==
The y–intercepts are 2 and 6.
21.
2
46xy y=− − +
This is a parabola of the form
2
xay byc=++.
Since 1a=− is negative, the parabola opens to the
Chapter 13 Review Exercises
Replace y with 0 to find the x–intercept.
()
22
04060066x=− − + = − + =
Solve using the quadratic formula.
The y–intercepts are 210−± .
22.
2
810xyy+=+
Since only one variable is squared, the graph of the
equation is a parabola.
23.
22
22
16 32
16 32
xy
xy
=−
+=
25.
22
22
4
4
xy
xy
=−
+=
26.
22
22
36 576 16
yx
=+
27.
()()
22
34
1
xy+−
+=
29.
22
55 180xy+=
Because
22
and xy have the same positive
coefficient, the equation’s graph is a circle.
94
Chapter 13 Conic Sections and Systems of Nonlinear Equations
Because the denominator of the
2
termx− is greater
than the denominator of the
2
term,y− the major
31.
22
49 36xy−=
Because
22
and xy have opposite signs, the
equation’s graph is a hyperbola.
The equation is in the form
22
22
1
xy
ab
−=
with
22
9, and 4ab==. We know the transverse axis
32.
22
1
25 1
xy
+=
Since only one variable is squared, the graph of the
equation is a parabola.
2
Chapter 13 Review Exercises
Solve using the quadratic formula.
()()
2
22413
y−± − − −
=−
34.
2
2
32
23
yxx
yx x
−= −
=−+
Since only one variable is squared, the graph of the
equation is a parabola.
This is a parabola of the form
2
yax bxc=++.
Since 1a= is positive, the parabola opens to the
right. The x–coordinate of the vertex is
The y–intercept is 3. Replace y with 0 to find the x–
intercepts.
2
023xx=−+
We do not need to simplify further. The solutions
are complex and there are no x–intercepts.
The center of the ellipse is
()
2, 5 .− Because the
denominator of the
2
termx− is greater than the
denominator of the
2
term,y− the major axis is
horizontal. Since
2
16,a= 4a= and the vertices
lie 4 units to the left and right of the center. Since
2
4,b= 2b= and endpoints of the minor axis lie
two units above and below the center.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
36.
()( )
22
324xy−++ =
22
37.
22
6260xy xy++−+=
Because
22
and xy have the same positive
coefficient, the equation’s graph is a circle.
()( )
22
22
6260
6 2 6
xy xy
xx yy
++−+=
++− =−
Complete the squares.
38. a. Using the point (6, 3), substitute for x and y to
find a in
2
.yax=
412
3
p
p
=
=
The light source should be placed at the point (0,
3). This is the point 3 inches above the vertex.
39.
2
51
1
yx
xy
=−
−=
41
xx
==
Back-substitute 1 and 4 for x to find y.
4or 1
11
41 11
30
xx
yx yx
yy
yy
==
=− =−
=− =−
==
The solution set is
()( )
{
}
1, 0 , 4, 3 .
Chapter 13 Review Exercises
40.
2
21
1
yx x
xy
=++
+=
Solve the second equation for y.
Back-substitute –3 and 0 for x to find y.
()
3
0or
1
1
01 3 1
131
4
x
x
yx
yx
yy
yy
y
=−
=
=− +
=− +
=− + =− − +
==+
=
The solution set is
()()
{
}
3, 4 , 0,1 .−
41.
22
2
0
xy
xy
+=
+=
Solve the second equation for y.
Back-substitute –1 and 1 for x to find y.
The solution set is
()
{
1,1 ,−
()
}
1, 1 .−
42.
22
22
224
15
xy
xy
+=
+=
Multiple the second equation by –1 and add to the
2
2
915
6
6
y
y
y
+=
=
=±
The solution set is
()
{
3, 6 ,−−
()()
3, 6 , 3, 6 ,−−
()
}
3, 6 .
43.
40
0
xy
yx
−=
−=
Solve the second equation for y.
0yx
−=
Back-substitute –2 and 2 for x to find y.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
44.
2
4
230
yx
xy
=
−+=
Substitute 2y – 3 for x in the first equation and
solve for y.
()
2
2
42 3
812
yy
yy
=−
=−
The solution set is
()( )
{
}
1, 2 , 9, 6 .
45.
22
10
2
xy
yx
+=
=+
Substitute x + 2 for y in the first equation and solve
for x.
Back-substitute –3 and 1 for x to find y.
{
}
46.
1
21
xy
yx
=
=+
()()
2
2
21
210
21 10
xx
xx
xx
+=
+−=
−+=
210 or 10
xx
−= +=
111
2
yy
y
=− =+
=
The solution set is
()
1
1, 1 , , 2
2
−−
.
47.
10
xy
++=
2
2
2555
xx
−−=−
2