MATH 1060 CA Ch 5 (Conics) Review
SOLUTIONS
The following are questions pertaining to the topics in College Algebra, Chapter 5, Conics, which is covered
at the University of Utah in Math 1060. For more study material, refer to past homework, quizzes, lecture
notes and problems in the textbook.
1. If the conic is not in standard form, rewrite it in this form. Then identify the conic, give the requested
information (see conic-specific parts below), and sketch its graph, labeling important points with their
coordinates.
• Circle: Center and radius
• Parabola: Vertex, Focus,Directrix, orientation
• Ellipse: Center, length of semi-major axis, length of semi-minor axis (or 4 points where the axes
intersect the ellipse), foci
• Hyperbola: center, vertices, orientation, slopes of asymptotes
(a)
x2 (y − 3)2
−
=1
9
16
Solution:
y
First of all, consider the untranslated graph of
x2
y2
the given graph.
−
= 1. It has these
9
16
features:
m=−
4
3
• vertices (x-intercepts) are (3, 0) and
(−3, 0),
6
C = (0, 3)
V = (−3, 3)
V = (3, 3)
2
Now, if you translate all the information right 0
and up 3, we get:
x
• center: (0, 3)
−6
• vertices: (3, 3) and (−3, 3),
• orientation: (doesn’t change)
• foci: (5, 3) and (−5, 3)
• slopes of asymptotes:
4
3
8
• orientation: horizontal hyperbola
• foci are (5, 0) and (−5, 0)
4
4
• and asymptotes are y = x and y = − x.
3
3
m=
10
4
−4
and
3
3
−4
−2
2
−2
−4
4
6
(b)
(x + 3)2 y 2
+
= 4.
9
16
Solution:
y
First, rewrite it in standard form by dividing
(x + 3)2
y2
both sides by 4:
+
= 1.
36
64
It is an ellipse with these features:
8
6
• vertical ellipse
4
• centered at (−3, 0)
• length of semi-major axis is 8 (this is the
axis parallel to the y-axis), length of semiminor axis is 6 (this is parallel to the xaxis)
• We find the foci:
2
4
C = (−3, 0)
−10
−8
−6
−4
x
−2
2
3
−2
62 + c2 = 82
−4
36 + c2 = 64
−6
c2 = 28
√
c=2 7
√
−8
√
foci are (−3, 2 7) and (−3, −2 7)
Page 2
4
(c) (y − 2)2 = −12(x − 3)
Solution:
y
It is
12
directix
x=6
10
8
• a horizontal parabola (concave left)
6
V =
(3, 2)
4
• vertex is at (3, 2)
F = (0, 2) 2
• Since 4c = 12, c = 3. Therefore, it has the
focus at (0, 2) and its directrix is x = 6.
−6 −4 −2
−2
2
4
6
x
−4
−6
(d) x2 + y 2 − 4y = 0
Solution:
y
First of all, let us complete the square.
x2 + (y 2 − 4y) = x2 + (y 2 − 4y + 4) − 4
2
4
2
= x + (y − 2) − 4.
2
Therefore, the given equation becomes
C=
(0, 2)
x2 + (y − 2)2 − 4 = 0
x
x2 + (y − 2)2 = 4.
−4
−2
2
It is a circle of radius 2 centered at (0, 2).
−2
Page 3
4
(e) −x2 + 16y 2 − 6x − 32y − 9 = 0
Solution:
First of all, let us complete the square.
−x2 + 16y 2 − 6x − 32y − 9 = 0
translate all the information left 3 and up 1, the
hyperbola we get has these features:
−(x2 + 6x) + 16(y 2 − 2y) − 9 = 0
• center: (−3, 1)
−(x2 + 6x + 9 − 9) + 16(y 2 − 2y + 1 − 1) − 9 = 0
• vertices: (−3, 2) and (−3, 0),
−(x2 + 6x + 9) + 9 + 16(y 2 − 2y + 1) − 16 − 9 = 0
−(x + 3)2 + 16(y − 1)2 = 16
• orientation:
(x + 3)2
−
+ (y − 1)2 = 1
16
or
(y − 1)2 −
1
1
and −
4
4
• slopes of asymptotes:
(x + 3)2
=1
16
y
4
V = (−3, 2)
It is a vertical hyperbola centered at (-3,1). To
plot the graph, consider the untranslated graph
of the given graph
2
C = (−3, 1)
x
x2
− + y 2 = 1.
16
−6
−4
−2
V = (−3, 0)
Then, it has vertices
√ (y-intercepts)√at (0, 1) and
(0, −1), foci at (0, 17) and (0, − 17) and two
1
1
asymptotes y = x and y = − x. Now, if you
4
4
2
4
−2
−4
(f) x2 − 6x + 8y − 31 = 0
Solution:
• vertex is at (3, 5)
First of all, let us complete the square for the
given equation.
• Since 4c = −8, c = −2. Therefore, the
focus is at (3, 3) and directix is y = 7.
x2 − 6x + 8y − 31 = x2 − 6x + 9 − 9 + 8y − 31
= (x − 3)2 + 8y − 40
y
Therefore, the equation becomes
(x − 3)2 + 8y − 40 = 0
6
V
4
F
2
(x − 3) = −8y + 40
(x − 3)2 = −8(y − 5)
2
It is
−6
−4
x
−2
2
−2
• a vertical parabola (concave down)
Page 4
4
6
8
10
2. Write the equation that matches the graph shown. use the equation formats given before Question 3
(a) The graph and its asymptotes are shown
y
8
Solution: Hyperbola opening
(y − k)2
(x − h)2
−
=1
2
a
b2
6
Center: (−1, 3)
4
(y − 3)2
(x + 1)2
−
=1
22
42
sideways:
2
−4
x
−2
2
4
−2
(b) Consider the graph below:
y
Solution: Ellipse:
2
−10
−8
−6
−4
x
−2
2
4
6
−2
(x − h)2
(y − k)2
+
=1
a2
b2
Center: (−2, −3)
Major axis: a = 6
Minor axis: b = 1
(x + 2)2
(y + 3)2
+
=1
2
6
12
−4
−6
(c) The graph and its latus rectum (focal chord) are shown
y
Solution: Parabola opening sideways in a
positive direction: (y − k)2 = 4c(x − h)
10
8
6
Vertex: (1, 4)
4c part: 4c = 4 so c = 1
4
(y − 4)2 = 4(x − 1)
2
x
−2
2
4
6
8
−2
Page 5
3. Use the information about each conic to find its equation in standard form and graph the equation.
Label the graph with the important information about the conic that you used to graph it.
(a) A circle with center (2, 4) that passes though a point (6, 6).
Solution:
Use the formula for a circle:
The equation is:
(x − 2)2 + (y − 4)2 = 20
(x − h)2 + (y − k)2 = r2
Insert the information about the vertex into the
formula: (h, k) = (2, 4) so
y
(x − 2)2 + (y − 4)2 = r2
Insert the information about the point (6, 6) into
the equation and solve for c:
8
(6, 6)
6
C = (2, 4)
2
2
(6 − 2) + (6 − 4) = r
4
2
42 + 2 2 = r 2
√
2
20 = r2
20 = r
√
2 5=r
x
−4
√
√
( 5 is a bit more than 2 so 2 5 is about 4.5)
Page 6
−2
2
−2
4
6
8
(b) A parabola has vertex (−1, 3), passes though a point (1, 5), and opens upward.
Solution:
Choose this formula because the vertex opens
upward:
The equation is:
(x − h)2 = 4c(y − k)
(x + 1)2 = 4
Insert the information about the vertex into the
formula: (h, k) = (−1, 3) so
1
(y − 3)
2
(x + 1)2 = 2(y − 3)
y
(x + 1)2 = 4c(y − 3)
Insert the information about the point (1, 5) into
the equation and solve for c:
(1 + 1)2 = 4c(5 − 3)
6
F 4
2
2 = 4c(2)
V
4 = 8c
1
=c
2
−6
−4
2
x
−2
2
4
(c) A hyperbola with vertices
at (4, −7) and (4, 3) and one focus at (4, −8).
√
(Use the fact that 11 ≈ 3.3.)
Solution:
By plotting the points, we can see the hyperbola
opens up-down, so 2we choose this
formula:
(x − h)2
(y − k)
−
=1
a2
b2
We find the midpoint between vertices: it has x−7 + 3
−4
coordinate 2 and y coordinate
=
=
2
2
−2. Thus the center is (4, −2)
Insert the information about the center and a
and b into the formula:
(y + 2)2
(x − 4)2
√
=1
−
52
( 11)2
y
6
We now find a, b, and c:
• the distance between the center and vertex: a = −2 − (−7) = −2 + 7 = 5
• the distance between the center and focus:
c = −2 − (−8) = −2 + 8 = 6
4
2
V = (4, 3)
x
−2
2
−2
4
6
8
C = (4, −2)
• Find b:
−4
2
2
c =a +b
62 = 52 + b2
36 = 25 + b2
11 = b
−6
V = (4, −7)
−8
F = (4, −8)
2
−10
2
Page 7
(d) A parabola has focus (0, 4) and directrix x = 10.
Solution:
First draw the information:
We insert the information about the vertex and
c into the formula:
x = 10
(y − 4)2 = 4(−5)(x − 5)
(y − 4)2 = −20(x − 5)
5
F = (0, 4)
5
y
16
V
directix
x = 10
14
12
10
Since the equation opens left, use this equation:
(y − k)2 = 4c(x − h).
8
6
The vertexis halfwaybetween the focus and the
0 + 10
, 4 = (5, 4)
directrix:
2
The distance from the focus the the vertex is c.
We solve for c:
F = (0, 4) 4
V =
(5, 4)
2
−6 −4 −2
−2
2
4
6
8
10 12
x
−4
c=0−5
−6
c = −5
(e) An ellipse with a vertical semi-major axis equal to 5, a horizontal semi-minor axis equal to 4,
and center (1, 1).
Solution:
y
Use the formula for a ellipse:
6
(x − h)2
(y − k)2
+
=1
2
a
b2
4
2
Insert the information about the center into the
formula: (h, k) = (1, 1) so
−2
2
−2
−4
Page 8
C = (1, 1)
4
−4
(x − 1)2
(y − 1)2
+
=1
42
52
5
x
4
6
4. Classify each conic as a circle, parabola, ellipse, or hyperbola and justify your reasoning. You can
assume none of the equations lead to degenerate conics.
(a) 3x2 + 2x − 3y 2 + 5y − 10 = 0
Solution: This is a hyperbola, because the coefficients of x2 and y 2 have different signs.
(b) x2 + x + 4y 2 − y − 1 = 0
Solution: This is an ellipse, because the coefficients of x2 and y 2 are both positive, but are different.
(c) x2 + 2x − 4y − 5 = 0
Solution: This is a parabola, because there is and x2 term and a y term, but there is no y 2 term.
Page 9