Module 7: Practice Problems
Conic Sections
About These Practice Problems
This set contains 20 comprehensive problems covering all topics in Module 7. Work through each problem carefully, then check your solution. Each problem includes detailed step-by-step explanations.
Problems 1-5: Circles
Problem 1: Write Equation in Standard Form
Write the equation of a circle with center (3, -2) and radius 5 in standard form.
Solution:
Step 1: Recall the standard form of a circle
(x - h)2 + (y - k)2 = r2
where (h, k) is the center and r is the radius.
(x - h)2 + (y - k)2 = r2
where (h, k) is the center and r is the radius.
Step 2: Identify the given values
Center: (h, k) = (3, -2)
Radius: r = 5
Center: (h, k) = (3, -2)
Radius: r = 5
Step 3: Substitute into the formula
(x - 3)2 + (y - (-2))2 = 52
(x - 3)2 + (y - (-2))2 = 52
Step 4: Simplify
(x - 3)2 + (y + 2)2 = 25
(x - 3)2 + (y + 2)2 = 25
Final Answer: (x - 3)2 + (y + 2)2 = 25
Problem 2: Find Center and Radius
Find the center and radius of the circle: (x + 4)2 + (y - 1)2 = 36
Solution:
Step 1: Identify the standard form
The equation is already in standard form: (x - h)2 + (y - k)2 = r2
The equation is already in standard form: (x - h)2 + (y - k)2 = r2
Step 2: Rewrite to match the pattern
(x + 4)2 = (x - (-4))2
(y - 1)2 = (y - 1)2
(x + 4)2 = (x - (-4))2
(y - 1)2 = (y - 1)2
Step 3: Identify h, k, and r
h = -4
k = 1
r2 = 36, so r = 6
h = -4
k = 1
r2 = 36, so r = 6
Final Answer: Center: (-4, 1), Radius: 6
Problem 3: Complete the Square
Write the equation x2 + y2 - 6x + 8y - 11 = 0 in standard form. Find the center and radius.
Solution:
Step 1: Group x and y terms
(x2 - 6x) + (y2 + 8y) = 11
(x2 - 6x) + (y2 + 8y) = 11
Step 2: Complete the square for x
x2 - 6x: Take half of -6 to get -3, then square it to get 9
(x2 - 6x + 9) = (x - 3)2
Add 9 to both sides
x2 - 6x: Take half of -6 to get -3, then square it to get 9
(x2 - 6x + 9) = (x - 3)2
Add 9 to both sides
Step 3: Complete the square for y
y2 + 8y: Take half of 8 to get 4, then square it to get 16
(y2 + 8y + 16) = (y + 4)2
Add 16 to both sides
y2 + 8y: Take half of 8 to get 4, then square it to get 16
(y2 + 8y + 16) = (y + 4)2
Add 16 to both sides
Step 4: Write the equation
(x - 3)2 + (y + 4)2 = 11 + 9 + 16
(x - 3)2 + (y + 4)2 = 36
(x - 3)2 + (y + 4)2 = 11 + 9 + 16
(x - 3)2 + (y + 4)2 = 36
Step 5: Identify center and radius
Center: (3, -4)
Radius: r = 6
Center: (3, -4)
Radius: r = 6
Final Answer: (x - 3)2 + (y + 4)2 = 36; Center: (3, -4), Radius: 6
Problem 4: Graph a Circle
Graph the circle (x - 2)2 + (y + 1)2 = 9. Identify the center and plot at least four key points.
Solution:
Step 1: Identify center and radius
Center: (h, k) = (2, -1)
Radius: r = 3 (since r2 = 9)
Center: (h, k) = (2, -1)
Radius: r = 3 (since r2 = 9)
Step 2: Plot the center
Plot the point (2, -1)
Plot the point (2, -1)
Step 3: Find four key points using the radius
Move 3 units in each direction from the center:
Right: (2 + 3, -1) = (5, -1)
Left: (2 - 3, -1) = (-1, -1)
Up: (2, -1 + 3) = (2, 2)
Down: (2, -1 - 3) = (2, -4)
Move 3 units in each direction from the center:
Right: (2 + 3, -1) = (5, -1)
Left: (2 - 3, -1) = (-1, -1)
Up: (2, -1 + 3) = (2, 2)
Down: (2, -1 - 3) = (2, -4)
Step 4: Draw the circle
Connect the four points with a smooth circular curve, maintaining equal distance from the center.
Connect the four points with a smooth circular curve, maintaining equal distance from the center.
Final Answer: Circle centered at (2, -1) with radius 3, passing through points (5, -1), (-1, -1), (2, 2), and (2, -4)
Problem 5: Equation from Endpoints of Diameter
Find the equation of a circle with diameter endpoints at (-2, 3) and (4, -1).
Solution:
Step 1: Find the center (midpoint of diameter)
Center = ((x1 + x2)/2, (y1 + y2)/2)
Center = ((-2 + 4)/2, (3 + (-1))/2)
Center = (2/2, 2/2) = (1, 1)
Center = ((x1 + x2)/2, (y1 + y2)/2)
Center = ((-2 + 4)/2, (3 + (-1))/2)
Center = (2/2, 2/2) = (1, 1)
Step 2: Find the radius (half the diameter)
First find the diameter using distance formula:
d = √[(4 - (-2))2 + (-1 - 3)2]
d = √[62 + (-4)2]
d = √[36 + 16] = √52 = 2√13
Radius = d/2 = √13
First find the diameter using distance formula:
d = √[(4 - (-2))2 + (-1 - 3)2]
d = √[62 + (-4)2]
d = √[36 + 16] = √52 = 2√13
Radius = d/2 = √13
Step 3: Write the equation
(x - 1)2 + (y - 1)2 = (√13)2
(x - 1)2 + (y - 1)2 = 13
(x - 1)2 + (y - 1)2 = (√13)2
(x - 1)2 + (y - 1)2 = 13
Final Answer: (x - 1)2 + (y - 1)2 = 13
Problems 6-9: Parabolas
Problem 6: Vertical Parabola - Find Vertex and Focus
For the parabola (x - 2)2 = 8(y + 1), find the vertex, focus, and directrix.
Solution:
Step 1: Identify the form
This is a vertical parabola in the form (x - h)2 = 4p(y - k)
This is a vertical parabola in the form (x - h)2 = 4p(y - k)
Step 2: Find h, k, and p
(x - 2)2 = 8(y - (-1))
h = 2, k = -1
4p = 8, so p = 2
(x - 2)2 = 8(y - (-1))
h = 2, k = -1
4p = 8, so p = 2
Step 3: Find the vertex
Vertex: (h, k) = (2, -1)
Vertex: (h, k) = (2, -1)
Step 4: Find the focus
For vertical parabola, focus is at (h, k + p)
Focus: (2, -1 + 2) = (2, 1)
For vertical parabola, focus is at (h, k + p)
Focus: (2, -1 + 2) = (2, 1)
Step 5: Find the directrix
Directrix: y = k - p = -1 - 2 = -3
Directrix: y = k - p = -1 - 2 = -3
Final Answer: Vertex: (2, -1), Focus: (2, 1), Directrix: y = -3
Problem 7: Horizontal Parabola - Write Equation
Write the equation of a horizontal parabola with vertex at (-3, 4) and focus at (1, 4).
Solution:
Step 1: Identify the form needed
Since vertex and focus have the same y-coordinate, this is a horizontal parabola: (y - k)2 = 4p(x - h)
Since vertex and focus have the same y-coordinate, this is a horizontal parabola: (y - k)2 = 4p(x - h)
Step 2: Find h and k from vertex
Vertex: (h, k) = (-3, 4)
h = -3, k = 4
Vertex: (h, k) = (-3, 4)
h = -3, k = 4
Step 3: Find p
Focus is at (h + p, k)
(1, 4) = (-3 + p, 4)
-3 + p = 1
p = 4
Focus is at (h + p, k)
(1, 4) = (-3 + p, 4)
-3 + p = 1
p = 4
Step 4: Write the equation
(y - 4)2 = 4(4)(x - (-3))
(y - 4)2 = 16(x + 3)
(y - 4)2 = 4(4)(x - (-3))
(y - 4)2 = 16(x + 3)
Final Answer: (y - 4)2 = 16(x + 3)
Problem 8: Convert to Standard Form
Write y2 - 4y - 12x + 16 = 0 in standard form and identify the vertex.
Solution:
Step 1: Isolate the squared variable term
y2 - 4y = 12x - 16
y2 - 4y = 12x - 16
Step 2: Complete the square for y
Take half of -4 to get -2, then square it to get 4
y2 - 4y + 4 = 12x - 16 + 4
(y - 2)2 = 12x - 12
Take half of -4 to get -2, then square it to get 4
y2 - 4y + 4 = 12x - 16 + 4
(y - 2)2 = 12x - 12
Step 3: Factor the right side
(y - 2)2 = 12(x - 1)
(y - 2)2 = 12(x - 1)
Step 4: Identify vertex
This is a horizontal parabola: (y - k)2 = 4p(x - h)
Vertex: (h, k) = (1, 2)
This is a horizontal parabola: (y - k)2 = 4p(x - h)
Vertex: (h, k) = (1, 2)
Final Answer: (y - 2)2 = 12(x - 1); Vertex: (1, 2)
Problem 9: Parabola from Focus and Directrix
Write the equation of a parabola with focus at (0, 3) and directrix y = -3.
Solution:
Step 1: Determine the orientation
Since the directrix is horizontal (y = -3), this is a vertical parabola.
Since the directrix is horizontal (y = -3), this is a vertical parabola.
Step 2: Find the vertex (midpoint between focus and directrix)
Vertex y-coordinate: (3 + (-3))/2 = 0
Vertex x-coordinate: same as focus = 0
Vertex: (0, 0)
Vertex y-coordinate: (3 + (-3))/2 = 0
Vertex x-coordinate: same as focus = 0
Vertex: (0, 0)
Step 3: Find p (distance from vertex to focus)
p = 3 - 0 = 3
p = 3 - 0 = 3
Step 4: Write the equation
Using (x - h)2 = 4p(y - k):
(x - 0)2 = 4(3)(y - 0)
x2 = 12y
Using (x - h)2 = 4p(y - k):
(x - 0)2 = 4(3)(y - 0)
x2 = 12y
Final Answer: x2 = 12y
Problems 10-14: Ellipses
Problem 10: Horizontal Ellipse - Find Vertices and Foci
For the ellipse (x - 1)2/25 + (y + 2)2/9 = 1, find the center, vertices, and foci.
Solution:
Step 1: Identify the center
Standard form: (x - h)2/a2 + (y - k)2/b2 = 1
Center: (h, k) = (1, -2)
Standard form: (x - h)2/a2 + (y - k)2/b2 = 1
Center: (h, k) = (1, -2)
Step 2: Identify a2 and b2
a2 = 25, so a = 5
b2 = 9, so b = 3
Since a > b, this is a horizontal ellipse (major axis is horizontal)
a2 = 25, so a = 5
b2 = 9, so b = 3
Since a > b, this is a horizontal ellipse (major axis is horizontal)
Step 3: Find the vertices
Major axis vertices: (h ± a, k) = (1 ± 5, -2)
Vertices: (-4, -2) and (6, -2)
Minor axis vertices: (h, k ± b) = (1, -2 ± 3)
Co-vertices: (1, 1) and (1, -5)
Major axis vertices: (h ± a, k) = (1 ± 5, -2)
Vertices: (-4, -2) and (6, -2)
Minor axis vertices: (h, k ± b) = (1, -2 ± 3)
Co-vertices: (1, 1) and (1, -5)
Step 4: Find c using a2 = b2 + c2
25 = 9 + c2
c2 = 16
c = 4
25 = 9 + c2
c2 = 16
c = 4
Step 5: Find the foci
Foci are on the major axis: (h ± c, k)
Foci: (1 - 4, -2) and (1 + 4, -2)
Foci: (-3, -2) and (5, -2)
Foci are on the major axis: (h ± c, k)
Foci: (1 - 4, -2) and (1 + 4, -2)
Foci: (-3, -2) and (5, -2)
Final Answer: Center: (1, -2), Vertices: (-4, -2) and (6, -2), Foci: (-3, -2) and (5, -2)
Problem 11: Write Equation of Ellipse
Write the equation of an ellipse centered at the origin with vertices at (0, 6) and (0, -6), and co-vertices at (4, 0) and (-4, 0).
Solution:
Step 1: Determine the orientation
Vertices are on the y-axis, so this is a vertical ellipse (major axis is vertical)
Vertices are on the y-axis, so this is a vertical ellipse (major axis is vertical)
Step 2: Find a and b
Distance from center to vertex: a = 6
Distance from center to co-vertex: b = 4
Distance from center to vertex: a = 6
Distance from center to co-vertex: b = 4
Step 3: Determine the form
For a vertical ellipse centered at origin: x2/b2 + y2/a2 = 1
For a vertical ellipse centered at origin: x2/b2 + y2/a2 = 1
Step 4: Write the equation
x2/16 + y2/36 = 1
x2/16 + y2/36 = 1
Final Answer: x2/16 + y2/36 = 1
Problem 12: Convert to Standard Form
Write 4x2 + 9y2 - 16x + 18y - 11 = 0 in standard form. Identify the center and lengths of the major and minor axes.
Solution:
Step 1: Group and factor
(4x2 - 16x) + (9y2 + 18y) = 11
4(x2 - 4x) + 9(y2 + 2y) = 11
(4x2 - 16x) + (9y2 + 18y) = 11
4(x2 - 4x) + 9(y2 + 2y) = 11
Step 2: Complete the square for x
x2 - 4x: Half of -4 is -2, squared is 4
4(x2 - 4x + 4) = 4(x - 2)2
Add 4(4) = 16 to the right side
x2 - 4x: Half of -4 is -2, squared is 4
4(x2 - 4x + 4) = 4(x - 2)2
Add 4(4) = 16 to the right side
Step 3: Complete the square for y
y2 + 2y: Half of 2 is 1, squared is 1
9(y2 + 2y + 1) = 9(y + 1)2
Add 9(1) = 9 to the right side
y2 + 2y: Half of 2 is 1, squared is 1
9(y2 + 2y + 1) = 9(y + 1)2
Add 9(1) = 9 to the right side
Step 4: Simplify and divide
4(x - 2)2 + 9(y + 1)2 = 11 + 16 + 9
4(x - 2)2 + 9(y + 1)2 = 36
(x - 2)2/9 + (y + 1)2/4 = 1
4(x - 2)2 + 9(y + 1)2 = 11 + 16 + 9
4(x - 2)2 + 9(y + 1)2 = 36
(x - 2)2/9 + (y + 1)2/4 = 1
Step 5: Identify key features
Center: (2, -1)
a2 = 9, so a = 3 (horizontal major axis)
b2 = 4, so b = 2
Major axis length: 2a = 6
Minor axis length: 2b = 4
Center: (2, -1)
a2 = 9, so a = 3 (horizontal major axis)
b2 = 4, so b = 2
Major axis length: 2a = 6
Minor axis length: 2b = 4
Final Answer: (x - 2)2/9 + (y + 1)2/4 = 1; Center: (2, -1), Major axis: 6, Minor axis: 4
Problem 13: Ellipse with Given Foci
Write the equation of an ellipse centered at the origin with foci at (0, 3) and (0, -3), and vertices at (0, 5) and (0, -5).
Solution:
Step 1: Identify the orientation
Foci and vertices are on the y-axis, so this is a vertical ellipse.
Foci and vertices are on the y-axis, so this is a vertical ellipse.
Step 2: Find a and c
Distance from center to vertex: a = 5
Distance from center to focus: c = 3
Distance from center to vertex: a = 5
Distance from center to focus: c = 3
Step 3: Find b using a2 = b2 + c2
25 = b2 + 9
b2 = 16
b = 4
25 = b2 + 9
b2 = 16
b = 4
Step 4: Write the equation
For vertical ellipse at origin: x2/b2 + y2/a2 = 1
x2/16 + y2/25 = 1
For vertical ellipse at origin: x2/b2 + y2/a2 = 1
x2/16 + y2/25 = 1
Final Answer: x2/16 + y2/25 = 1
Problem 14: Find Eccentricity
Find the eccentricity of the ellipse x2/36 + y2/20 = 1.
Solution:
Step 1: Identify a2 and b2
a2 = 36, so a = 6
b2 = 20, so b = √20 = 2√5
Since a > b, the major axis is horizontal
a2 = 36, so a = 6
b2 = 20, so b = √20 = 2√5
Since a > b, the major axis is horizontal
Step 2: Find c using a2 = b2 + c2
36 = 20 + c2
c2 = 16
c = 4
36 = 20 + c2
c2 = 16
c = 4
Step 3: Calculate eccentricity
Eccentricity: e = c/a
e = 4/6 = 2/3
Eccentricity: e = c/a
e = 4/6 = 2/3
For an ellipse, 0 < e < 1. The closer e is to 0, the more circular the ellipse. The closer to 1, the more elongated.
Final Answer: e = 2/3
Problems 15-20: Hyperbolas
Problem 15: Horizontal Hyperbola - Find Key Features
For the hyperbola (x - 2)2/16 - (y + 1)2/9 = 1, find the center, vertices, foci, and asymptotes.
Solution:
Step 1: Identify the center and orientation
Standard form: (x - h)2/a2 - (y - k)2/b2 = 1 (horizontal)
Center: (h, k) = (2, -1)
Standard form: (x - h)2/a2 - (y - k)2/b2 = 1 (horizontal)
Center: (h, k) = (2, -1)
Step 2: Find a and b
a2 = 16, so a = 4
b2 = 9, so b = 3
a2 = 16, so a = 4
b2 = 9, so b = 3
Step 3: Find the vertices
For horizontal hyperbola: (h ± a, k)
Vertices: (2 ± 4, -1)
Vertices: (-2, -1) and (6, -1)
For horizontal hyperbola: (h ± a, k)
Vertices: (2 ± 4, -1)
Vertices: (-2, -1) and (6, -1)
Step 4: Find c using c2 = a2 + b2
c2 = 16 + 9 = 25
c = 5
c2 = 16 + 9 = 25
c = 5
Step 5: Find the foci
For horizontal hyperbola: (h ± c, k)
Foci: (2 ± 5, -1)
Foci: (-3, -1) and (7, -1)
For horizontal hyperbola: (h ± c, k)
Foci: (2 ± 5, -1)
Foci: (-3, -1) and (7, -1)
Step 6: Find the asymptotes
For horizontal hyperbola: y - k = ±(b/a)(x - h)
y - (-1) = ±(3/4)(x - 2)
y + 1 = ±(3/4)(x - 2)
y = -1 + (3/4)(x - 2) and y = -1 - (3/4)(x - 2)
y = (3/4)x - 5/2 and y = -(3/4)x + 1/2
For horizontal hyperbola: y - k = ±(b/a)(x - h)
y - (-1) = ±(3/4)(x - 2)
y + 1 = ±(3/4)(x - 2)
y = -1 + (3/4)(x - 2) and y = -1 - (3/4)(x - 2)
y = (3/4)x - 5/2 and y = -(3/4)x + 1/2
Final Answer: Center: (2, -1), Vertices: (-2, -1) and (6, -1), Foci: (-3, -1) and (7, -1), Asymptotes: y = (3/4)x - 5/2 and y = -(3/4)x + 1/2
Problem 16: Write Equation of Hyperbola
Write the equation of a hyperbola centered at the origin with vertices at (0, 3) and (0, -3), and foci at (0, 5) and (0, -5).
Solution:
Step 1: Determine the orientation
Vertices and foci are on the y-axis, so this is a vertical hyperbola.
Vertices and foci are on the y-axis, so this is a vertical hyperbola.
Step 2: Find a and c
Distance from center to vertex: a = 3
Distance from center to focus: c = 5
Distance from center to vertex: a = 3
Distance from center to focus: c = 5
Step 3: Find b using c2 = a2 + b2
25 = 9 + b2
b2 = 16
b = 4
25 = 9 + b2
b2 = 16
b = 4
Step 4: Write the equation
For vertical hyperbola at origin: y2/a2 - x2/b2 = 1
y2/9 - x2/16 = 1
For vertical hyperbola at origin: y2/a2 - x2/b2 = 1
y2/9 - x2/16 = 1
Final Answer: y2/9 - x2/16 = 1
Problem 17: Convert to Standard Form
Write 9x2 - 4y2 - 18x - 16y - 43 = 0 in standard form and identify the center.
Solution:
Step 1: Group and factor
(9x2 - 18x) - (4y2 + 16y) = 43
9(x2 - 2x) - 4(y2 + 4y) = 43
(9x2 - 18x) - (4y2 + 16y) = 43
9(x2 - 2x) - 4(y2 + 4y) = 43
Step 2: Complete the square for x
x2 - 2x: Half of -2 is -1, squared is 1
9(x2 - 2x + 1) = 9(x - 1)2
Add 9(1) = 9 to the right side
x2 - 2x: Half of -2 is -1, squared is 1
9(x2 - 2x + 1) = 9(x - 1)2
Add 9(1) = 9 to the right side
Step 3: Complete the square for y
y2 + 4y: Half of 4 is 2, squared is 4
-4(y2 + 4y + 4) = -4(y + 2)2
Add -4(4) = -16 to the right side
y2 + 4y: Half of 4 is 2, squared is 4
-4(y2 + 4y + 4) = -4(y + 2)2
Add -4(4) = -16 to the right side
Step 4: Simplify and divide
9(x - 1)2 - 4(y + 2)2 = 43 + 9 - 16
9(x - 1)2 - 4(y + 2)2 = 36
(x - 1)2/4 - (y + 2)2/9 = 1
9(x - 1)2 - 4(y + 2)2 = 43 + 9 - 16
9(x - 1)2 - 4(y + 2)2 = 36
(x - 1)2/4 - (y + 2)2/9 = 1
Step 5: Identify center
Center: (1, -2)
Center: (1, -2)
Final Answer: (x - 1)2/4 - (y + 2)2/9 = 1; Center: (1, -2)
Problem 18: Find Asymptotes
Find the equations of the asymptotes for the hyperbola y2/25 - x2/16 = 1.
Solution:
Step 1: Identify the form and center
This is a vertical hyperbola centered at the origin: y2/a2 - x2/b2 = 1
Center: (0, 0)
This is a vertical hyperbola centered at the origin: y2/a2 - x2/b2 = 1
Center: (0, 0)
Step 2: Find a and b
a2 = 25, so a = 5
b2 = 16, so b = 4
a2 = 25, so a = 5
b2 = 16, so b = 4
Step 3: Use the asymptote formula for vertical hyperbola
For vertical hyperbola: y - k = ±(a/b)(x - h)
Since center is at origin: y = ±(a/b)x
y = ±(5/4)x
For vertical hyperbola: y - k = ±(a/b)(x - h)
Since center is at origin: y = ±(a/b)x
y = ±(5/4)x
Final Answer: y = (5/4)x and y = -(5/4)x
Problem 19: Identify the Conic Section
Identify the conic section represented by the equation 4x2 + 4y2 - 8x + 16y - 20 = 0.
Solution:
Step 1: Check the coefficients of x2 and y2
Coefficient of x2: 4
Coefficient of y2: 4
They are equal and have the same sign.
Coefficient of x2: 4
Coefficient of y2: 4
They are equal and have the same sign.
Step 2: Apply the identification rule
When coefficients of x2 and y2 are equal and have the same sign, the conic is a circle.
When coefficients of x2 and y2 are equal and have the same sign, the conic is a circle.
Step 3: Verify by converting to standard form (optional)
4(x2 - 2x) + 4(y2 + 4y) = 20
4(x2 - 2x + 1) + 4(y2 + 4y + 4) = 20 + 4 + 16
4(x - 1)2 + 4(y + 2)2 = 40
(x - 1)2 + (y + 2)2 = 10
This is indeed a circle.
4(x2 - 2x) + 4(y2 + 4y) = 20
4(x2 - 2x + 1) + 4(y2 + 4y + 4) = 20 + 4 + 16
4(x - 1)2 + 4(y + 2)2 = 40
(x - 1)2 + (y + 2)2 = 10
This is indeed a circle.
Final Answer: Circle
Problem 20: Application - Satellite Dish
A satellite dish has a parabolic cross-section. The dish is 16 feet across and 4 feet deep. If the receiver is placed at the focus, how far from the vertex should it be placed?
Solution:
Step 1: Set up a coordinate system
Place the vertex at the origin with the parabola opening upward.
Equation form: x2 = 4py
Place the vertex at the origin with the parabola opening upward.
Equation form: x2 = 4py
Step 2: Use the given dimensions
The dish is 16 feet across and 4 feet deep.
At the edge of the dish: x = 8 (half of 16 feet) and y = 4
The dish is 16 feet across and 4 feet deep.
At the edge of the dish: x = 8 (half of 16 feet) and y = 4
Step 3: Substitute into the equation
82 = 4p(4)
64 = 16p
p = 4
82 = 4p(4)
64 = 16p
p = 4
Step 4: Interpret p
The value p = 4 represents the distance from the vertex to the focus.
The value p = 4 represents the distance from the vertex to the focus.
The receiver should be placed at the focus because all incoming parallel rays (satellite signals) will reflect off the parabolic surface and converge at the focus.
Final Answer: The receiver should be placed 4 feet from the vertex.
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