For each equation, identify the vertex, focus, axis of symmetry, and directrix. Then graph the parabola.
1.(x – 3)2 = 12(y – 7)
SOLUTION:
The equation is in standard form and the squared term is x, which means that the parabola opens vertically. The
equation is in the form (x − h)2 = 4p(y − k), so h = 3 and k = 7. Because 4p = 12 and p = 3, the graph opens
up.Use the values of h, k, and p to determine the characteristics of the parabola.
vertex: (h, k) = (3, 7)
focus: (h, k + p) = (3, 10)
directrix: y = k – p or 4
axis of symmetry: x = h or 3
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape
of the curve.
3.(y – 4)2 = 20(x + 2)
SOLUTION:
The equation is in standard form and the squared term is y,whichmeansthattheparabolaopenshorizontally.The
equation is in the form (y − k)2 = 4p(x − h), so h = –2 and k = 4. Because 4p = 20 and p = 5, the graph opens to
the right. Use the values of h, k, and p to determine the characteristics of the parabola.
vertex: (h, k) = (–2, 4)
focus: (h + p, k) = (3, 4)
directrix: x = h – p or –7
axisofsymmetry:y = k or 4
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape
of the curve.
5.(x + 8)2 = 8(y – 3)
SOLUTION:
The equation is in standard form and the squared term is x, which means that the parabola opens vertically. The
equation is in the form (x − h)2 = 4p(y − k), so h = –8 and k = 3. Because 4p = 8 and p = 2, the graph opens up.
Use the values of h, k, and p to determine the characteristics of the parabola.
vertex: (h, k) = (–8, 3)
focus: (h, k + p) = (–8, 5)
directrix: y = k – p or 1
axisofsymmetry:x = h or –8
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape
of the curve.
7.(y + 5)2 = 24(x − 1)
SOLUTION:
The equation is in standard form and the squared term is y, which means that the parabola opens horizontally. The
equation is in the form (y − k)2 = 4p(x − h), so h = 1 and k = –5. Because 4p = 24 and p = 6, the graph opens to
the right. Use the values of h, k, and p to determine the characteristics of the parabola.
vertex: (h, k) = (1, –5)
focus: (h + p, k) = (7, –5)
directrix: x = h – p or –5
axisofsymmetry:y = k or –5
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape
of the curve.
9.−4(y +2) = (x + 8)2
SOLUTION:
The equation is in standard form and the squared term is x, which means that the parabola opens vertically. The
equation is in the form 4p(y − k) = (x − h)2, so h = –8 and k = –2. Because 4p = –4 and p = –1, the graph opens
down. Use the values of h, k, and p to determine the characteristics of the parabola.
vertex: (h, k) = (–8, –2)
focus: (h + p, k) = (–8, –3)
directrix: y = k – p or –1
axisofsymmetry:x = h or –8
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape
of the curve.
11.SKATEBOARDING A group of high school students designing a half–pipe have decided that the ramps, or
transitions, could be obtained by splitting a parabola in half. A parabolic cross section of the ramps can be modeled
by x2 = 8(y − 2), where the values of x and y are measured in feet. Where is the focus of the parabola in relation
to the ground if the ground represents the directrix?
SOLUTION:
Because the x–term is squared and p is positive, the parabola opens up and the focus is located at (h, k + p). The
equation is provided in standard form, and h = 0 and k = 2. Because 4p = 8, p is 2. So, the location of the focus is
(0, 2 + 2) or (0, 4). Therefore, the focus is 4 feet above the ground.
13.BOATINGAsaspeedboatglidesthroughthewater,itcreatesawakeintheshapeofaparabola.Thevertexof
this parabola meets with the stern of the boat. A swimmer on a wakeboard, attached by a piece of rope, is being
pulled by the boat. When he is directly behind the boat, he is positioned at the focus of the parabola. The parabola
formed by the wake can be modeled using y2 − 180x + 10y + 565 = 0, where x and y are measured in feet.
a. Write the equation in standard form.
b. How long is the length of rope attaching the swimmer to the stern of the boat?
SOLUTION:
a.
b.The equation in standard form has y as the squared term, which means that the parabola opens horizontally.
Because 4p = 180, p = 45 and the graph opens to the right. The equation is in the form (y − k)2 = 4p(x − h), so h =
3 and k = –5. Since the stern is located at the vertex of the parabola formed, it is at the point (h, k) or (3, –5).
The swimmer is at the focus, located at (h + p, k), which is (3 + 45, –5) or (48, –5). The distance the swimmer is
fromthesternoftheboatrepresentsthelengthofropeneededtoattachtheswimmertothestern.
Using the distance formula, the distance between these two points is or45.
Therefore, the length of rope attaching the swimmer to the stern of the boat is 45 feet.
Write each equation in standard form. Identify the vertex, focus, axis of symmetry, and directrix. Then
graph the parabola.
15.x2 − 17 = 8y + 39
SOLUTION:
Because the x–term is squared and p = 2, the graph opens up. Use the standard form to determine the
characteristics of the parabola.
vertex: (h, k) = (0, –7)
focus: (h, k + p) = (0, –5)
directrix: y = k – p or –9
axisofsymmetry:x = h or 0
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the curve. The
curve should be symmetric about the axis of symmetry.
17.3x2 + 72 = −72y
SOLUTION:
Because the x–term is squared and p = −6, the graph opens down. Use the standard form to determine the
characteristics of the parabola.
vertex: (h, k) = (0, –1)
focus: (h, k + p) = (0, –7)
directrix: y = k – p or 5
axis of symmetry: x = h or 0
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the curve. The
curve should be symmetric about the axis of symmetry.
19.60x – 80 = 3y2 + 100
SOLUTION:
Because the y–term is squared and p = 5, the graph opens to the right. Use the standard form to determine the
characteristics of the parabola.
vertex: (h, k) = (3, 0)
focus: (h + p, k) = (8, 0)
directrix: x = h – p or −2
axis of symmetry: y = k or 0
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the curve. The
curve should be symmetric about the axis of symmetry.
21.−72 = 2y2 – 16y − 20x
SOLUTION:
Because the y–term is squared and p = 2.5, the graph opens to the right. Use the standard form to determine the
characteristics of the parabola.
vertex: (h, k) = (2, 4)
focus: (h + p, k) = (4.5, 4)
directrix: x = h – p or −0.5
axis of symmetry: y = k or 4
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the curve. The
curve should be symmetric about the axis of symmetry.
23.x2 – 18y + 12x = 126
SOLUTION:
Because the x–term is squared and p = 4.5, the graph opens up. Use the standard form to determine the
characteristics of the parabola.
vertex: (h, k) = (−6, −9)
focus: (h, k + p) = (−6, −4.5)
directrix: y = k – p or −13.5
axis of symmetry: x = h or –6
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the curve. The
curve should be symmetric about the axis of symmetry.
25.LIGHTING Stadium lights at an athletic field need to reflect light at maximum intensity. The bulb should be placed
at the focal point of the parabolic globe surrounding it. If the shape of the globe is given by x2 = 36y, where x and y
are in inches, how far from the vertex of the globe should the bulb be placed for maximum light?
SOLUTION:
The bulb is placed at the focus of the parabola. Because the x–term is squared and p is positive, the parabola
opens up and the focus is located at (h, k + p). The equation is provided in standard form, and h = 0 and k = 0.
Because 4p = 36, p is 9. So, the location of the focus is (0, 0 + 9) or (0, 9). The vertex of the globe is located at (0,
0). The distance from the focus to the vertex is 9 − 0 or 9. Therefore, the bulb should be placed 9 inches from the
vertex of the globe.
Write an equation for and graph a parabola with the given focus F and vertex V.
27.F(2, −1), V(−4, −1)
SOLUTION:
Because the focus and vertex share the same y–coordinate, the graph is horizontal. The focus is (h + p, k), so the
value of p is 2 – (−4) or 6. Because p is positive, the graph opens to the right.
Write the equation for the parabola in standard form using the values of h, p, and k.
The standard form of the equation is (y + 1)2 = 24(x + 4).Graph the vertex and focus. Then make a table of values
to graph the parabola.
4p(x – h) = (y – k)2
4(6)[x – (−4)] = [y – (−1)]2
24(x + 4) = (y + 1)2
29.F(−3, 4), V(−3, 2)
SOLUTION:
Because the focus and vertex share the same x–coordinate, the graph is vertical. The focus is (h, k + p), so the
value of p is 4 − 2 or 2. Because p is positive, the graph opens up.
Write the equation for the parabola in standard form using the values of h, p, and k.
The standard form of the equation is (x + 3)2 = 8(y – 2). Graph the vertex and focus. Then make a table of values
to graph the parabola.
4p(y – k) = (x – h)2
4(2)(y – 2) = [x − (−3)]2
8(y − 2) = (x + 3)2
31.F(−1, 4), V(7, 4)
SOLUTION:
Because the focus and vertex share the same y–coordinate, the graph is horizontal. The focus is (h + p, k), so the
value of p is −1 − 7 or −8. Because p is negative, the graph opens to the left.
Write the equation for the parabola in standard form using the values of h, p, and k.
The standard form of the equation is (y – 4)2 = −32(x – 7). Graph the vertex and focus. Then make a table of
values to graph the parabola.
4p(x – h) = (y – k)2
4(−8)(x − 7) = (y – 4)2
−32(x − 7) = (y − 4)2
33.F(1, 3), V(1, 6)
SOLUTION:
Because the focus and vertex share the same x–coordinate, the graph is vertical. The focus is (h, k + p), so the
value of p is 3 − 6 or −3. Because p is negative, the graph opens down.
Write the equation for the parabola in standard form using the values of h, p, and k.
The standard form of the equation is (x – 1)2 = −12(y – 6). Graph the vertex and focus. Then make a table of
values to graph the parabola.
4p(y – k) = (x – h)2
4(−3)(y – 6) = (x – 1)2
−12(y − 6) = (x − 1)2
35.F(8, −3), V(8, −7)
SOLUTION:
Because the focus and vertex share the same x–coordinate, the graph is vertical. The focus is (h, k + p), so the
value of p is −3 − (−7) or 4. Because p is positive, the graph opens up.
Write the equation for the parabola in standard form using the values of h, p, and k.
The standard form of the equation is (x – 8)2 = 16(y + 7). Graph the vertex and focus. Then make a table of values
to graph the parabola.
4p(y – k) = (x – h)2
4(4)[y – (−7)] = (x – 8)2
16(y + 7) = (x − 8)2
Write an equation for and graph each parabola with focus F and the given characteristics.
37.F(1, 2); opens down; contains (7, 2)
SOLUTION:
Because the parabola opens down, the vertex is (1, 2 − p). Use the standard form of the equation of a vertical
parabola and the point (7, 2) to find the equation.
Because the parabola opens down, the value of p must be negative. Therefore, p = −3. The vertex is (1, 5), and the
standard form of the equation is (x – 1)2 = −12(y − 5). Use a table of values to graph the parabola.
4p(y – k) = (x – h)2
4p[2 – (2 − p)] = (7 – 1)2
4p(p) = 36
p2 = 9
p = ±3
39.F(−4, 0); opens down; contains (4, −15)
SOLUTION:
Because the parabola opens down, the vertex is (−4, −p). Use the standard form of the equation of a vertical
parabola and the point (4, −15) to find the equation.
Because the parabola opens down, the value of p must be negative. Therefore, p = −1. The vertex is (−4, 1), and
the standard form of the equation is (x + 4)2 = −4(y − 1). Use a table of values to graph the parabola.
4p(y – k) = (x – h)2
4p[−15 – (−p)] = [4 – (−4)]2
4p(−15 + p) = 64
4p2 − 60p = 64
p2 − 15p − 16 = 0
(p − 16)(p + 1) = 0
p = 16 or −1
41.F(−5, −9); opens right; contains (10, −1)
SOLUTION:
Because the parabola opens to the right, the vertex is (−5 − p, −9). Use the standard form of the equation of a
horizontal parabola and the point (10, −1) to find the equation.
Because the parabola opens to the right, the value of p must be positive. Therefore, p = 1. The vertex is (−6, −9),
and the standard form of the equation is 4(x + 6) = (y + 9)2. Use a table of values to graph the parabola.
4p(x – h) = (y – k)2
4p[10 – (−5 − p)] = [−1 – (−9)]2
4p(15 + p) = 64
4p2 + 60p = 64
p2 + 15p − 16 = 0
(p + 16)(p − 1) = 0
p = 1 or −16
43.F(−5, −2); opens up; contains (−13, −2)
SOLUTION:
Because the parabola opens up, the vertex is (−5, −2 − p). Use the standard form of the equation of a horizontal
parabola and the point (−13, −2) to find the equation.
Because the parabola opens up, the value of p must be positive. Therefore, p = 4. The vertex is (−5, −6), and the
standard form of the equation is (x + 5)2 = 16(y + 6). Use a table of values to graph the parabola.
4p(y – k) = (x – h)2
4p[−2 – (−2 − p)] = [−13 – (−5)]2
4p(p) = 64
p2 = 16
p = ±4
Write an equation for the line tangent to each parabola at each given point.
45.
SOLUTION:
The graph opens vertically. Determine the vertex and focus. iswritteninstandardform.
Because . The vertex is (−7, 3) and the focus is (−7, 2.875). We need to determine d, the
distance between the focus and the point of tangency, C.
This is one leg of the isosceles triangle.
Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is negative, the parabola opens down
and A will be above the focus.
Points A and C both lie on the line tangent to the parabola. Find an equation of this line.
47.(x + 6)2 = 3(y – 2), (0, 14)
SOLUTION:
The graph opens vertically. Determine the vertex and focus. (x + 6)2 = 3(y − 2) is written in standard form.
Because 4p = 3, p = . The vertex is (−6, 2) and the focus is (−6, 2.75). We need to determine d, the distance
between the focus and the point of tangency, C.
This is one leg of the isosceles triangle.
Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is positive, the parabola opens up and
A will be below the focus.
A = (−6, 2.75 − 12.75) or (−6, −10)
Points A and C both lie on the line tangent to the parabola. Find an equation of this line.
d =
=
= 12.75
m = or4
y – y1 = m(x – x1)
y – 14 = 4(x – 0)
y − 14 = 4x
y = 4x + 14
49.−0.25(x – 6)2 = y – 9, (10, 5)
SOLUTION:
The graph opens vertically. Determine the vertex and focus.
Because 4p = −4, p = −1. The vertex is (6, 9) and the focus is (6, 8). We need to determine d, the distance
between the focus and the point of tangency, C.
This is one leg of the isosceles triangle.
Use d to find A, the endpoint of the other leg of the isosceles triangle. Since p is negative, the parabola opens down
and A will be above the focus.
A = (6, 8 + 5) or (6, 13)
Points A and C both lie on the line tangent to the parabola. Find an equation of this line.
−0.25(x − 6)2 = y − 9
(x – 6)2 = −4(y – 9)
d =
=
= 5
m = or −2
y – y1 = m(x – x1)
y – 5 = −2(x – 10)
y − 5 = −2x + 20
y = −2x + 25
Determine the orientation of each parabola.
51.directrix y = 4, p = −2
SOLUTION:
Since the directrix is y = 4, the graph opens vertically. Since p is negative, the graph will open down.
53.vertex (−5, 1), focus (−5, 3)
SOLUTION:
Because the focus and vertex share the same x–coordinate, the graph is vertical. The focus is (h, k + p), so the
value of p is 3 − 1or 2. Because p is positive, the graph opens up.
Write an equation for each parabola.
55.
SOLUTION:
The focus and vertex share the same x–coordinate. The focus is (h, k + p), so the value of p is 4 − 5 or −1.
Write the equation for the parabola in standard form using the values of h, p, and k.
4p(y – k) = (x – h)2
4(−1)(y – 5) = (x – 3)2
−4(y − 5) = (x − 3)2
57.
SOLUTION:
The focus is (−9, 1) and the directrix is x = −1. The directrix is x = h − p or h − p = −1 and the x–coordinate of the
focus is h + p = −9.
Solve the system of equations.
Substitute h = −5 into the equation for the directrix to find that p is –4.
Write the equation for the parabola in standard form using the values of h, p, and k.
h − p
h + p = −1
= −9
2h = −10
h = −5
4p(x – h) = (y – k)2
4(−4)[x – (−5)] = (y – 1)2
−16(x + 5) = (y − 1)2
59.BRIDGESThearchoftherailroadtrackbridgebelowisintheshapeofaparabola.Thetwomainsupport
towers are 208 meters apart and 80 meters tall. The distance from the top of the parabola to the water below is 60
meters.
a. Write an equation that models the shape of the arch. Let the railroad track represent the x–axis.
b. Two vertical supports attached to the arch are equidistant from the center of the parabola as shown in the
diagram. Find their lengths if they are 86.4 meters apart.
SOLUTION:
a. If the railroad track represents the x–axis and we assume that the vertex lies on the y–axis, then the vertex is at
the point (0, −20). The arch meets each support tower 104 meters to the left and to the right of the vertex and 80
feet below the railroad or the x–axis. Thus, two points on the parabola are at (−104, −80) and (104, −80). Find p
using the vertex and a point on the parabola.
Write the equation for the parabola in standard form using the values of h, p, and k.
b. If the supports are 86.4 meters apart and are equidistant from the center of the parabola, then they are located
43.2 meters from the center of the parabola. Therefore, the supports meet the parabola when x = 43.2. Find y
when x = 43.2.
Therefore, the supports meets the parabola at the points (43.2, −30.35) and (−43.2, −30.35) and the supports are
30.35 meters in length.
4p(y – k) = (x – h)2
4p [−80 – (−20)] = (104 – 0)2
−240p = 10816
p = −45.066
4p(y – k) = (x – h)2
4(−45.066)[y – (−20)] = (x – 0)2
−180.27(y + 20) = x2
−180.27(y + 20) = x2
−180.27(y + 20) = 43.22
−180.27(y + 20) = 1866.24
y + 20 = −10.35
y = −30.35
Write an equation for and graph a parabola with each set of characteristics.
61.vertex (−6, 4), contains (−10, 8), opens horizontally
SOLUTION:
If the graph opens horizontally, then the standard form is 4p(x − h) = (y − k)2. Find p using the vertex and the point
on the parabola.
Write the equation for the parabola in standard form using the values of h, p, and k.
4p(x – h) = (y – k)2
4p[−10 – (−6)] = (8 – 4)2
−16p = 16
p = −1
4p(x – h) = (y – k)2
4(−1)[x – (−6)] = (y – 4)2
−4(x + 6) = (y − 4)2
63.opens horizontally, passes through points (−1, −1), (5, 3), and (15, 7)
SOLUTION:
If the graph opens horizontally, then the standard form is 4p(x − h) = (y − k)2. Find an equation for the parabola for
each set of points.
Equation 1:
Equation 2:
Equation 3:
Solve this system of equations. Subtract Equation 2 from Equation 1.
Subtract Equation 2 from Equation 3.
Solve the system consisting of the two new equations.
Substitute p = 2 into one of the equations and solve for k.
Substitute p = 2 and k = −5 into one of the original equations and solve for h.
Write the equation for the parabola in standard form using the values of h, p, and k.
4p(x – h) = (y – k)2
4p (−1 – h) = (−1 – k)2
−4p − 4hp = 1 + 2k + k2
4p(x – h) = (y – k)2
4p(5 – h) = (3 – k)2
20p − 4hp = 9 − 6k + k2
4p(x – h) = (y – k)2
4p (15 – h) = (7 – k)2
60p − 4hp = 49 − 14k + k2
−4p − 4hp = 1 + 2k + k2
20p − 4hp = 9 − 6k + k2
−24p = −8 + 8k
60p − 4hp = 49 − 14k + k2
20p − 4hp = 9 − 6k + k2
40p = 40 − 8k
−24p
40p = −8 + 8k
= 40 − 8k
16p = 32
p = 2
40p = 40 − 8k
40(2) = 40 − 8k
80 = 40 − 8k
40 = −8k
−5 = k
−4p − 4hp = 1 + 2k + k2
−4(2) − 4(2)h = 1 + 2(−5) + (−5)2
−8 − 8h = 16
−8h = 24
h = −3
4p(x – h) = (y – k)2
4(2)[x – (−3)] = [y – (−5)]2
8(x + 3) = (y + 5)2
Determine the point of tangency for each equation and line.
65.(x + 2)2 = 2y
y = 4x
SOLUTION:
The point of tangency occurs at the solution for the system of equations.
Use substitution to solve the system.
The solution to the system occurs when x = 2. To find the y–coordinate, substitute x = 2 in one of the original
equations.
(x + 2)2 = 2y
(x + 2)2 = 2(4x)
x2 + 4x + 4 = 8x
x2 − 4x + 4 = 0
(x − 2)2 = 0
y = 4x
y = 4(2)
y = 8
The point of tangency occurs at (2, 8).
67.(y + 3)2 = −x + 4
SOLUTION:
The point of tangency occurs at the solution for the system of equations.
Use substitution to solve the system.
The solution to the system occurs when x = 0. To find the y–coordinate, substitute x = 0 into one of the original
equations.
y = −1
The point of tangency occurs at (0, −1).
(y + 3)2 = −x + 4
= −x + 4
= −x + 4
= −x + 4
= 0
69.ILLUMINATIONIn a searchlight, the bulb is placed at the focus of a parabolic mirror 1.5 feet from the vertex.
This causes the light rays from the bulb to bounce off the mirror as parallel rays, thus providing a concentrated
beam of light.
a.Write an equation for the parabola if the focal diameter of the bulb is 2 feet, as shown in the diagram.
b.Suppose the focal diameter is increased to 3 feet. If the depth of both searchlights is 3.5 feet, how much greater
is the width of the opening of the larger light? Round to the nearest hundredth.
SOLUTION:
a. Sketch a coordinate plane on top of the searchlight with the vertex at the origin. The focus is 1.5 feet from the
vertex located at the point (1.5, 0). Since the focal diameter of the bulb is 2 feet, 1 foot will lie above and below the
x–axis. Therefore, the points (1.5, 1) and (1.5, −1) lie on the parabola. Substitute the values of x, y, h, and k to
solve for p.
Substitute values for h, k, and p to write an equation to model the reflector.
b. Write an equation in standard form for a parabola with a focal diameter of 3 feet. Sketch a coordinate plane on
top of the new searchlight with the vertex at the origin. The focus is still 1.5 feet from the vertex located at the
point (1.5, 0). Since the focal diameter of the bulb is 3 feet, 1.5 feet will lie above and below the x–axis. Therefore,
the points (1.5, 1.5) and (1.5, –1.5) lie on the parabola. Substitute the values of x, y, h, and k to solve for p.
Substitute values for h, k, and p to write an equation to model the reflector.
The depth of both searchlights is 3.5 feet. The depth represents the distance the reflector spans horizontally and the
width represents the distance the reflector spans vertically. To find the width of the first searchlight, substitute x =
3.5 into the equation found in part a and solve for y.
At a depth of 3.5 feet, the reflector will span 1.53 feet above and below the x-axis for a total width of 2(1.53) or
3.06 feet. Repeat the same process to find the width of the second reflector.
At a depth of 3.5 feet, the reflector will span 2.29 feet above and below the x-axis for a total width of 2(2.29) or
4.58 feet. The difference between the two widths is 4.58 – 3.06 or 1.52 feet.
4p(x − h) = (y − k)2
4p(1.5 − 0) = (1 − 0)2
6p = 1
p =
4p(x − h) = (y − k)2
4 (x − 0) = (y − 0)2
x = y2
4p(x − h) = (y − k)2
4p(1.5 − 0) = (1.5 − 0)2
6p = 2.25
p = 0.375
4p(x − h) = (y − k)2
4(0.375)(x − 0) = (y − 0)2
1.5x = y2
x = y2
(3.5) = y2
= y2
± = y
± or ±1.53 = y
1.5x = y2
1.5(3.5) = y2
5.25 = y2
± or 2.29 = y
71.The latus rectum of a parabola is the line segment that passes through the focus, is perpendicular to the axis of
the parabola, and has endpoints on the parabola. The length of the latus rectum is |4p| units, where p is the distance
from the vertex to the focus.
a.Write an equation for the parabola with vertex at (−3, 2), axis y = 2, and latus rectum 8 units long.
b.Prove that the endpoints of the latus rectum and point of intersection of the axis and directrix are the vertices of
a right isosceles triangle.
SOLUTION:
a. Since the parabola has an axis of y = 2, the parabola will open horizontally and will have the standard form of (y
– k)2 = 4p(x – h). The latus rectum is 8 units long. This can be used to solve for p.
|4p| = 8, so p = ±2.
Substitute values for h, k, and pto write an equation for the parabola.
b.
To prove that the endpoints of the latus rectum X and W and the point of intersection of the axis and directrix D are
the vertices of a right isosceles triangle ∆XDW, we need to show that XDW is a right angle and that .
Since , , and XFD WFD, ∆XFD ∆FWD by SAS. Thus, . To prove that
XDW is a right angle, we can first prove that XDF and FDW are both angles.Since and
XFD is a right angle, ∆XFD is an isosceles right triangle. Therefore, XDF is . This also means that
FDW is . Resultantly, XDW is a right angle. Thus, ∆XDW is an isosceles right triangle.
4p(x − h) = (y − k)2
4(±2)[x − (−3)] = (y − 2)2
±8(x + 3) = (y − 2)2
Write a possible equation for a parabola with focus F such that the line given is tangent to the parabola.
73.
SOLUTION:
Find point A, the point of intersection of the axis and the line tangent to the parabola.
The equation for the tangent line is y = 2x − 2. Since point A lies on the y–axis, the x–coordinate is 0. Substitute x =
0 to find y.
y = 2x − 2
y = 2(0) − 2
y = −2
A is at (0, −2). The segment formed by FA represents one of the legs of the isosceles triangle formed by F, A, and
the point of tangency. Find the length of this leg.
d = 5
The lengths of the legs of the isosceles triangle are 5 units. This is the distance from F to the point of tangency (x,
y). Use the distance formula and the tangent line to solve for the point of tangency.
The solutions to this equation are x = 0 and 4. Since A has an x–coordinate of 0, the point of tangency must have
an x–coordinate of 4. Substitute x = 4 to solve for y.
y = 2x − 2
y = 2(4) − 2
y = 6
The point of tangency is located at (4, 6). Since the parabola opens upward, the standard form of the equation is (x
− h)2 = 4p(y − k) and the focus is located at (h, k + p). The focus is at the point (0, 3); thus, h = 0 and k + p = 3 or
k = 3 − p. Use, h, the standard form of the equation, the point of tangency, and the equation for k to solve for p.
The solutions to this equation are p = –4 and 1. Since the parabola opens upward, p is positive; thus, p = 1. Use p
to find k. k = 3 − (1) or 2. The vertex of the parabola is (0, 2). Use the vertex and p to write the standard form of
the equation.
d
5
25 = x2 + (y − 3)2
25 = x2 + y2 − 6y + 9
25 = x2 + (2x − 2)2 − 6(2x − 2) + 9
25 = x2 + 4x2 − 8x + 4 − 12x + 12 + 9
25 = 5x2 − 20x + 25
0 = 5x2 − 20x
0 = 5x(x − 4)
(x − h)2 = 4p(y − k)
(4 − 0)2 = 4p(6 − k)
16 = 24p − 4pk
16 = 24p − 4p(3 − p)
16 = 24p − 12p + 4p2
0 = 4p2 + 12p − 16
0 = 4(p + 4)(p − 1)
(x − h)2 = 4p(y − k)
(x − 0)2 = 4(1)(y − 2)
x2 = 4(y − 2)
75.
SOLUTION:
Find point A, the point of intersection of the axis and the line tangent to the parabola.
The equation for the tangent line is y = x + 6. Since point A lies on the x–axis, the y–coordinate is 0. Substitute y
= 0 to find x.
y = x + 6
0 = x + 6
−6 = x
−12 = x
A is at (−12, 0). The segment formed by FA represents one of the legs of the isosceles triangle formed by F, A, and
the point of tangency. Find the length of this leg.
d =
d =
d = 15
The lengths of the legs of the isosceles triangle are 15 units. This is the distance from F to the point of tangency (x,
y). Use the distance formula and the tangent line to solve for the point of tangency.
The solutions to this equation are x = ±12. Since A has an x–coordinate of −12, the point of tangency must have an
x–coordinate of 12. Substitute x = 12 to solve for y.
y = x + 6
y = (12) + 6
y = 12
The point of tangency is located at (12, 12). Since the parabola opens to the right, the standard form of the equation
is 4p(x − h) = (y − k)2 and the focus is located at (h + p, k). The focus is at the point (3, 0); thus, k = 0 and h + p
= 3 or h = 3 − p. Use, k, the standard form of the equation, the point of tangency, and the equation for h to solve
for p.
The solutions to this equation are p = −12 and 3. Since the parabola opens to the right, p is positive; thus, p = 3.
Use p to find h. h = 3 − (3) or 0. The vertex of the parabola is (0, 0). Use the vertex and p to write the standard
form of the equation.
d =
15 =
225 = (x − 3)2 + y2
225 = x2 −6x + 9 + y2
225 = x2− 6x + 9 +
225 = x2 − 6x + 9 + x2 + 6x + 36
225 = x2 + 45
180 = x2
144 = x2
±12 = x
4p(x − h) = (y − k)2
4p(12 − h) = (12 − 0)2
48p − 4ph = 144
48p − 4p(3 − p) = 144
48p − 12p + 4p2 = 144
4p2 + 36p − 144 = 0
4(p + 12)(p − 3) = 0
4p(x − h) = (y − k)2
4(3)(x − 0) = (y − 0)2
12x = y2
77.MULTIPLEREPRESENTATIONS In this problem, you will investigate how the shape of a parabola changes
as the position of the focus changes.
a. GEOMETRICFindthedistancebetweenthevertexandthefocusofeachparabola.
i. y2 = 4(x – 2)
ii. y2 = 8(x – 2)
iii. y2 = 16(x – 2)
b. GRAPHICALGraphtheparabolasinparta using a different color for each. Label each focus.
c. VERBALDescribetherelationshipbetweena parabola's shape and the distance between its vertex and focus.
d. ANALYTICALWriteanequationforaparabolathathasthesamevertexas(x + 1)2 = 20(y + 7), but is
narrower.
e. ANALYTICALMakeaconjectureaboutthegraphsofx2 = −2(y + 1), x2 = −12(y + 1), and x2 = −5(y + 1).
Check your conjecture by graphing the parabolas.
SOLUTION:
a.
i. Written in standard form, (y − k)2 = 4p(x − h), p is the distance between the vertex and the focus. y2 = 4(x − 2)
can be written as y2 = 4(1)(x − 2), where p = 1. Thus, the distance between the vertex and the focus is 1 unit.
ii. Written in standard form, (y − k)2 = 4p(x − h), p is the distance between the vertex and the focus. y2 = 8(x − 2)
can be written as y2 = 4(2)(x − 2), where p = 2. Thus, the distance between the vertex and the focus is 2 units.
iii. Written in standard form, (y − k)2 = 4p(x − h), p is the distance between the vertex and the focus. y2 = 16(x −
2) can be written as y2 = 4(4)(x − 2), where p = 4. Thus, the distance between the vertex and the focus is 4 units.
b. Use a table of values to graph each parabola.
c. As the focus is moved farther away from the vertex, the parabola becomes wider.
d. The parabola (x + 1)2 = 20(y + 7) can be written as (x + 1)2 = (4)(5)(y + 7), where p = 5 and the vertex is (−1,
−7). A parabola that is more narrow has a smaller value for p. Let p = 1 and write the standard form for the
equation of a parabola with the same vertex.
e. Sample answer: The parabolas all have a vertex of (0, −1) and open downward. The first equation produces the
most narrow parabola and the second equation produces the widest parabola.
(x − h)2 = 4p(y − k)
[x − (−1)]2 = 4(1)[y − (−7)]
(x + 1)2 = 4(y + 7)
79.CHALLENGE The area of the parabolic sector shaded in the graph below is given by
A = xy.Findtheequationoftheparabolaifthesectorareais2.4squareunitsandthewidthofthesectoris3
units.
SOLUTION:
The width of the sector is 3 units, so y = 1.5. A = xy and the area of the sector is 2.4 square units. Substitute the
values into the equation and solve for x.
2.4 = x(1.5)
2.4 = 2x
1.2 = x
Therefore, the vertex of the parabola is (0, 0), and two other points on the parabola are (1.2, 1.5) and (1.2, −1.5).
Set up and solve a system of equations using y as the independent variable.
The equation of the parabola is y2 = x or y2 = x.
ay2 + by + c = x
a(0)2 + b(0) + c = 0
a(1.5)2 + b(1.5) + c = 1.2
a(−1.5)2 + b(−1.5) + c = 1.2
c = 0
2.25a + 1.5b = 1.2
2.25a − 1.5b = 1.2
4.5a = 2.4
a =
2.25· +1.5b = 1.2
1.2 + 1.5b = 1.2
1.5b = 0
b = 0
81.REASONING Without graphing, determine in which quadrants the graph of (y – 5)2 = −8(x + 2) will have no
points. Explain your reasoning.
SOLUTION:
Quadrants I and IV; the vertex is (−2, 5) and p = −2. Since the vertex is to the left of the y–axis, and the parabola
opens to the left, no points will be to the right of the y–axis, or in Quadrants I and IV.
83.PREWRITE Write a letter outlining and explaining the content you have learned from this lesson. Make an
outline that addresses purpose, audience, a controlling idea, logical sequence, and time frame for completion.
SOLUTION:
Sample answer:
Purpose: To outline and explain the process of graphing parabolas.
Audience: Parents of PreCalculus students.
Controlling idea: Using characteristics of parabolas to graph them.
Logical sequence:
Find the vertex of the parabola.
Find the focus of the parabola.
Find the directrix of the parabola.
Find the axis of symmetry of the parabola.
Graph those characteristics.
Make a table of values and graph the general shape of the curve.
Timeframe: You have one day to complete the outline.
Find the maximum and minimum values of the objective function f(x, y) and for what values of x and y
they occur, subject to the given constraints.
85.y ≥ −x + 2
2 ≤ x≤ 7
y ≤ x + 5
f(x, y) = 8x + 3y
SOLUTION:
Begin by graphing the given system of three inequalities. The solution of the system, which makes up the set of
feasible solutions for the objective function, is the shaded region, including its boundary segments.
The polygonal region of feasible solutions has four vertices.
Solve the four systems of two boundary equations below to find the coordinates of the remaining vertices.
To solve the first system, substitute x = 2 into the boundary equation and solve for y.
y = x + 5
y = (2) + 5
y = 6
A vertex is located at (2, 6).
To solve the second system, substitute x = 7 into the boundary equation and solve for y.
y = x + 5
y = (7) + 5
y = 8.5
A vertex is located at (7, 8.5).
To solve the third system, substitute x = 2 into the boundary equation and solve for y.
y = −x + 2
y = −(2) + 2
y = 0
A vertex is located at (2, 0).
To solve the fourth system, substitute x = 7 into the boundary equation and solve for y.
y = −x + 2
y = −(7) + 2
y = −5
A vertex is located at (7, −5).
Find the value of the objective function f(x, y) = 8x + 3y at each of the four vertices.
So, the maximum value of f is 81.5 when x = 7 and y = 8.5. The minimum value of f is 16 when x = 2 and y = 0.
System of Boundary
Equations y = x + 5
x = 2
y = x + 5
x = 7
System of Boundary
Equations y = −x + 2
x = 2 y = −x + 2
x = 7
f(2, 6) = 8(2) + 3(6) or 34
f(7, 8.5) = 8(7) + 3(8.5) or 81.5
f(2, 0) = 8(2) + 3(0) or 16
f(7, −5) = 8(7) + 3(−5) or 41
87.Findthepartialfractiondecompositionof .
SOLUTION:
Rewrite the equation as partial fractions with constant numerators, A and B, and denominators that are the linear
factors of the original denominator.
= +
2y + 5 = A(y + 1) + B(y + 2)
2y + 5 = Ay + A + By + 2B
2y + 5 = (A + B)y + (A + 2B)
Equate the coefficients on the left and right side of the equation to obtain a system of two equations.
To solve the system, let A = 2 − B and use substitution.
Thus, A = 2 − (3) or −1. So, the partial fraction decomposition of .
A + B = 2
A + 2B = 5
(2 − B) + 2B = 5
2 + B = 5
B = 3
Find the value of each expression using the given information.
89.cot θ and csc θ; tan θ = , sec θ > 0
SOLUTION:
Use the reciprocal identity to find cot θ.
cot θ =
cot θ =
cot θ =
Use the Pythagorean identity and cot θ to find csc θ.
Since sec θ > 0 and cot is positive, csc > 0. Thus, csc = .
cot2 θ + 1 = csc2 θ
+ 1 = csc2 θ
+ = csc2 θ
= csc2 θ
± = csc θ
± = csc θ
Locate the vertical asymptotes, and sketch the graph of each function.
91.y = tan x + 4
SOLUTION:
The graph of y = tan θ +4 is the graph of y = tan x shifted 4 units up. The period is or π. Find the location of
two consecutive vertical asymptotes.
and
Create a table listing the coordinates of key points for y = tan θ +4 for one period on .
Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch more to the
left and right of the first curve.
Function y = tan x y = tan x +
4
Vertical
Asymptote
Intermediate
Point
x-intercept (0, 0) (0, 4)
Intermediate
Point
Vertical
Asymptote
93.y = csc x −
SOLUTION:
The graph of y = csc x − is the graph of y = csc x shifted unitdown.Theperiodis or π. Find the
location of two consecutive vertical asymptotes.
and
Create a table listing the coordinates of key points for y = csc x – for one period on .
Sketch the curve through the indicated key points for the function. Then repeat the pattern to sketch more to the
left and right of the first curve.
Function y = csc x y = csc x –
Vertical
Asymptote x = −π x = −π
Intermediate
Point
Vertical
Asymptote x = 0
Intermediate
Point
Vertical
Asymptote x = π x = π
Evaluate each expression.
95.log4 16x
SOLUTION:
log4 16x = log4 (42)x
= log4 42x
= 2x
Graph each function.
97.
f(x) = x3 – x2 – 4x + 4
SOLUTION:
Use a table of values to graph the function.
x f(x)
−3 −20
−2 0
−1 6
0 4
1 0
2 0
3 10
99.h(x) = x4 – 4x2 + 2
SOLUTION:
Use a table of values to graph the function.
x f(x)
−3 47
−2 2
−1 −1
0 2
1 −1
2 2
3 47
101.SAT/ACT If x is a positive number, then
F
G
H
J
SOLUTION:
=
=
=
103.REVIEW What are the x–intercepts of the graph of y = −2x2 – 5x + 12?
F
G −4,
H −2,
J
SOLUTION:
The x–intercepts of a graph occur at the zeros of the function. Factor to solve 0 = −2x2 − 5x + 12 for x.
0 = −2x2 − 5x + 12
0 = (−2x + 3)(x + 4)
The solutions to this equation occur when either of the factors is equal to 0.
The x–intercepts of the graph of y are −4 and .
−2x + 3 = 0
−2x = −3
x =
x + 4 = 0
x = −4
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7-1 Parabolas
