Lesson 2: Parabolas
Learning Objectives
- Understand the focus-directrix definition of a parabola
- Identify the vertex, focus, directrix, and axis of symmetry from an equation
- Write equations of parabolas in standard form (vertical and horizontal)
- Graph parabolas that open up, down, left, or right
- Determine the direction a parabola opens based on the value of p
- Convert between standard form and general form of parabola equations
- Find the equation of a parabola given key features (vertex, focus, or directrix)
- Apply parabolas to real-world problems involving satellite dishes, bridges, and projectile motion
Introduction: What is a Parabola?
Parabola: The set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix).
You have encountered parabolas before when studying quadratic functions like y = x². However, parabolas are actually one of the four conic sections (circles, parabolas, ellipses, and hyperbolas), and they have a beautiful geometric definition based on distance.
Key Components of a Parabola:
- Vertex: The point on the parabola closest to the directrix (the turning point)
- Focus: A fixed point inside the parabola
- Directrix: A fixed line outside the parabola
- Axis of Symmetry: The line through the vertex and focus, perpendicular to the directrix
- p: The directed distance from the vertex to the focus (also equals the distance from vertex to directrix)
The parameter p determines how "wide" or "narrow" the parabola is and which direction it opens. The sign of p indicates the direction:
- For vertical parabolas: p > 0 opens upward, p < 0 opens downward
- For horizontal parabolas: p > 0 opens rightward, p < 0 opens leftward
Standard Form: Vertical Parabolas
When a parabola opens upward or downward (vertical parabola), its equation has the form:
Where:
- Vertex: (h, k)
- Focus: (h, k + p)
- Directrix: y = k - p
- Axis of Symmetry: x = h (vertical line through vertex)
- If p > 0, the parabola opens upward
- If p < 0, the parabola opens downward
Example 1: Identifying Features of a Vertical Parabola
Problem: Find the vertex, focus, directrix, and axis of symmetry for the parabola (x - 2)² = 8(y + 1). Then determine which direction it opens.
Solution:
Step 1: Identify h, k, and p from the standard form (x - h)² = 4p(y - k)
Comparing (x - 2)² = 8(y + 1) with (x - h)² = 4p(y - k):
- h = 2
- k = -1 (note: y + 1 = y - (-1))
- 4p = 8, so p = 2
Step 2: Find the vertex
Vertex = (h, k) = (2, -1)
Step 3: Find the focus
Focus = (h, k + p) = (2, -1 + 2) = (2, 1)
Step 4: Find the directrix
Directrix: y = k - p = -1 - 2 = -3
Step 5: Find the axis of symmetry
Axis of symmetry: x = h = 2
Step 6: Determine direction
Since p = 2 > 0, the parabola opens upward.
Answer: Vertex (2, -1), Focus (2, 1), Directrix y = -3, Axis of symmetry x = 2, Opens upward
Example 2: Writing the Equation of a Vertical Parabola
Problem: Write the equation in standard form for a parabola with vertex (3, 4) and focus (3, 6).
Solution:
Step 1: Determine if the parabola is vertical or horizontal
Both points have the same x-coordinate (3), so this is a vertical parabola.
Use the form: (x - h)² = 4p(y - k)
Step 2: Identify h and k from the vertex
Vertex = (h, k) = (3, 4), so h = 3, k = 4
Step 3: Find p
The focus is at (h, k + p) = (3, 6)
So k + p = 6
4 + p = 6
p = 2
Step 4: Calculate 4p
4p = 4(2) = 8
Step 5: Write the equation
(x - 3)² = 8(y - 4)
Answer: (x - 3)² = 8(y - 4)
Example 3: Vertical Parabola Opening Downward
Problem: Find the equation of a parabola with vertex (-1, 5) and directrix y = 8.
Solution:
Step 1: Identify h and k
Vertex = (h, k) = (-1, 5), so h = -1, k = 5
Step 2: Find p using the directrix
Directrix: y = k - p
8 = 5 - p
p = 5 - 8 = -3
Note: p is negative, so the parabola opens downward
Step 3: Calculate 4p
4p = 4(-3) = -12
Step 4: Write the equation
(x - (-1))² = -12(y - 5)
(x + 1)² = -12(y - 5)
Answer: (x + 1)² = -12(y - 5)
Verification: Focus at (-1, 5 + (-3)) = (-1, 2), which is below the vertex, confirming downward opening
Example 4: Graphing a Vertical Parabola
Problem: Graph the parabola x² = 12y. Identify the vertex, focus, and directrix.
Solution:
Step 1: Rewrite in standard form
x² = 12y can be written as (x - 0)² = 12(y - 0)
So h = 0, k = 0, and 4p = 12, giving p = 3
Step 2: Find key features
- Vertex: (0, 0)
- Focus: (0, 3)
- Directrix: y = -3
- Axis of symmetry: x = 0 (the y-axis)
- Opens upward (p > 0)
Step 3: Plot points
Use the equation x² = 12y to find points:
- When x = 0: y = 0 (vertex)
- When x = 6: 36 = 12y, so y = 3
- When x = -6: 36 = 12y, so y = 3
- When x = 12: 144 = 12y, so y = 12
- When x = -12: 144 = 12y, so y = 12
Step 4: Sketch the graph
Plot the vertex at (0, 0), mark the focus at (0, 3), draw the directrix y = -3, and sketch the U-shaped parabola through the points, opening upward.
Answer: Vertex (0, 0), Focus (0, 3), Directrix y = -3
Example 5: Finding Vertex from Standard Form
Problem: For the parabola (x + 4)² = -20(y - 2), find the vertex and determine the direction it opens.
Solution:
Step 1: Rewrite to match standard form
(x + 4)² = -20(y - 2) is the same as (x - (-4))² = -20(y - 2)
So h = -4, k = 2
Step 2: Find the vertex
Vertex = (h, k) = (-4, 2)
Step 3: Determine p and direction
4p = -20, so p = -5
Since p < 0, the parabola opens downward
Answer: Vertex (-4, 2), Opens downward
Standard Form: Horizontal Parabolas
When a parabola opens left or right (horizontal parabola), its equation has the form:
Where:
- Vertex: (h, k)
- Focus: (h + p, k)
- Directrix: x = h - p
- Axis of Symmetry: y = k (horizontal line through vertex)
- If p > 0, the parabola opens to the right
- If p < 0, the parabola opens to the left
Important Note: Horizontal parabolas are NOT functions because they fail the vertical line test. However, they are still valid parabolas and important in applications.
Example 6: Identifying Features of a Horizontal Parabola
Problem: Find the vertex, focus, directrix, and axis of symmetry for the parabola (y - 3)² = 16(x + 2).
Solution:
Step 1: Identify h, k, and p from the standard form (y - k)² = 4p(x - h)
Comparing (y - 3)² = 16(x + 2) with (y - k)² = 4p(x - h):
- k = 3
- h = -2 (note: x + 2 = x - (-2))
- 4p = 16, so p = 4
Step 2: Find the vertex
Vertex = (h, k) = (-2, 3)
Step 3: Find the focus
Focus = (h + p, k) = (-2 + 4, 3) = (2, 3)
Step 4: Find the directrix
Directrix: x = h - p = -2 - 4 = -6
Step 5: Find the axis of symmetry
Axis of symmetry: y = k = 3
Step 6: Determine direction
Since p = 4 > 0, the parabola opens to the right.
Answer: Vertex (-2, 3), Focus (2, 3), Directrix x = -6, Axis of symmetry y = 3, Opens right
Example 7: Writing the Equation of a Horizontal Parabola
Problem: Write the equation for a parabola with vertex (1, -2) and focus (-3, -2).
Solution:
Step 1: Determine if the parabola is vertical or horizontal
Both points have the same y-coordinate (-2), so this is a horizontal parabola.
Use the form: (y - k)² = 4p(x - h)
Step 2: Identify h and k from the vertex
Vertex = (h, k) = (1, -2), so h = 1, k = -2
Step 3: Find p
The focus is at (h + p, k) = (-3, -2)
So h + p = -3
1 + p = -3
p = -4
Note: p is negative, so the parabola opens to the left
Step 4: Calculate 4p
4p = 4(-4) = -16
Step 5: Write the equation
(y - (-2))² = -16(x - 1)
(y + 2)² = -16(x - 1)
Answer: (y + 2)² = -16(x - 1)
Example 8: Horizontal Parabola from Directrix
Problem: Find the equation of a parabola with vertex (0, 0) and directrix x = -5.
Solution:
Step 1: Determine orientation
The directrix is a vertical line (x = -5), so this is a horizontal parabola.
Step 2: Identify h and k
Vertex = (h, k) = (0, 0), so h = 0, k = 0
Step 3: Find p using the directrix
Directrix: x = h - p
-5 = 0 - p
p = 5
Note: p is positive, so the parabola opens to the right (away from the directrix)
Step 4: Calculate 4p
4p = 4(5) = 20
Step 5: Write the equation
(y - 0)² = 20(x - 0)
y² = 20x
Answer: y² = 20x
Verification: Focus at (0 + 5, 0) = (5, 0), which is to the right of the vertex
Example 9: Graphing a Horizontal Parabola
Problem: Graph the parabola y² = -8x. Identify the vertex, focus, and directrix.
Solution:
Step 1: Rewrite in standard form
y² = -8x can be written as (y - 0)² = -8(x - 0)
So h = 0, k = 0, and 4p = -8, giving p = -2
Step 2: Find key features
- Vertex: (0, 0)
- Focus: (-2, 0)
- Directrix: x = 2
- Axis of symmetry: y = 0 (the x-axis)
- Opens to the left (p < 0)
Step 3: Plot points
Use the equation y² = -8x to find points (solve for x = -y²/8):
- When y = 0: x = 0 (vertex)
- When y = 4: 16 = -8x, so x = -2
- When y = -4: 16 = -8x, so x = -2
- When y = 8: 64 = -8x, so x = -8
- When y = -8: 64 = -8x, so x = -8
Step 4: Sketch the graph
Plot the vertex at (0, 0), mark the focus at (-2, 0), draw the directrix x = 2, and sketch the sideways parabola opening to the left.
Answer: Vertex (0, 0), Focus (-2, 0), Directrix x = 2
Example 10: Comparing Vertical and Horizontal Forms
Problem: Determine whether each parabola is vertical or horizontal, then find the vertex:
(a) (x - 5)² = 24(y + 1)
(b) (y + 3)² = -12(x - 4)
Solution:
Part (a): (x - 5)² = 24(y + 1)
The squared term is (x - h)², so this is a vertical parabola.
Form: (x - h)² = 4p(y - k)
h = 5, k = -1
Vertex: (5, -1)
Part (b): (y + 3)² = -12(x - 4)
The squared term is (y - k)², so this is a horizontal parabola.
Form: (y - k)² = 4p(x - h)
k = -3, h = 4
Vertex: (4, -3)
Answer: (a) Vertical, vertex (5, -1); (b) Horizontal, vertex (4, -3)
Summary Table: Vertical vs. Horizontal Parabolas
| Feature | Vertical Parabola | Horizontal Parabola |
|---|---|---|
| Standard Form | (x - h)² = 4p(y - k) | (y - k)² = 4p(x - h) |
| Squared Variable | x is squared | y is squared |
| Vertex | (h, k) | (h, k) |
| Focus | (h, k + p) | (h + p, k) |
| Directrix | y = k - p | x = h - p |
| Axis of Symmetry | x = h (vertical line) | y = k (horizontal line) |
| Opens Up/Right (p > 0) | Opens upward | Opens rightward |
| Opens Down/Left (p < 0) | Opens downward | Opens leftward |
| Is it a function? | Yes | No |
Converting Between Standard and General Forms
Sometimes parabolas are given in general form (expanded form), and we need to convert to standard form by completing the square.
Example 11: Converting to Standard Form (Vertical)
Problem: Convert x² - 6x - 8y + 1 = 0 to standard form and identify the vertex.
Solution:
Step 1: Isolate the terms with y on one side
x² - 6x + 1 = 8y
Step 2: Complete the square on x
Take half of -6, square it: (-6/2)² = 9
Add 9 to both sides:
x² - 6x + 9 + 1 = 8y + 9
(x - 3)² + 1 = 8y + 9
(x - 3)² = 8y + 8
Step 3: Factor the right side
(x - 3)² = 8(y + 1)
Step 4: Identify the vertex
Form: (x - h)² = 4p(y - k)
h = 3, k = -1
Vertex: (3, -1)
Also: 4p = 8, so p = 2 (opens upward)
Answer: Standard form: (x - 3)² = 8(y + 1); Vertex: (3, -1)
Example 12: Converting to Standard Form (Horizontal)
Problem: Convert y² + 4y + 12x - 8 = 0 to standard form and identify the vertex and focus.
Solution:
Step 1: Isolate the terms with x on one side
y² + 4y - 8 = -12x
Step 2: Complete the square on y
Take half of 4, square it: (4/2)² = 4
Add 4 to both sides:
y² + 4y + 4 - 8 = -12x + 4
(y + 2)² - 8 = -12x + 4
(y + 2)² = -12x + 12
Step 3: Factor the right side
(y + 2)² = -12(x - 1)
Step 4: Identify the vertex and focus
Form: (y - k)² = 4p(x - h)
h = 1, k = -2
Vertex: (1, -2)
4p = -12, so p = -3
Focus: (h + p, k) = (1 + (-3), -2) = (-2, -2)
Answer: Standard form: (y + 2)² = -12(x - 1); Vertex: (1, -2); Focus: (-2, -2)
Example 13: Complete Conversion with All Features
Problem: For the parabola x² + 8x + 4y + 12 = 0, convert to standard form and find the vertex, focus, directrix, and direction of opening.
Solution:
Step 1: Isolate the y term
x² + 8x + 12 = -4y
Divide by -4:
-1/4(x² + 8x + 12) = y
Or equivalently: x² + 8x + 12 = -4y
Step 2: Complete the square on x
Take half of 8, square it: (8/2)² = 16
Add 16 to both sides:
x² + 8x + 16 + 12 = -4y + 16
(x + 4)² + 12 = -4y + 16
(x + 4)² = -4y + 4
(x + 4)² = -4(y - 1)
Step 3: Identify h, k, and p
Form: (x - h)² = 4p(y - k)
h = -4, k = 1
4p = -4, so p = -1
Step 4: Find all features
- Vertex: (-4, 1)
- Focus: (h, k + p) = (-4, 1 + (-1)) = (-4, 0)
- Directrix: y = k - p = 1 - (-1) = 2
- Axis of symmetry: x = -4
- Opens downward (p < 0)
Answer: Standard form: (x + 4)² = -4(y - 1); Vertex: (-4, 1); Focus: (-4, 0); Directrix: y = 2; Opens downward
Finding Equations from Given Information
Example 14: Parabola Through Three Points
Problem: Find the equation of the vertical parabola with vertex at the origin that passes through the point (6, 3).
Solution:
Step 1: Write the general form for a vertical parabola with vertex at origin
(x - 0)² = 4p(y - 0)
x² = 4py
Step 2: Substitute the point (6, 3)
6² = 4p(3)
36 = 12p
p = 3
Step 3: Write the equation
x² = 4(3)y
x² = 12y
Answer: x² = 12y
Verification: When x = 6: 36 = 12y, so y = 3
Example 15: Using Focus and Directrix
Problem: Find the equation of the parabola with focus at (2, 5) and directrix y = 1.
Solution:
Step 1: Determine orientation
The directrix is horizontal (y = 1), so this is a vertical parabola.
Step 2: Find the vertex
The vertex is midway between the focus and directrix.
The focus is at (2, 5) and the directrix is at y = 1.
Vertex y-coordinate: k = (5 + 1)/2 = 3
Vertex x-coordinate: h = 2 (same as focus)
Vertex: (2, 3)
Step 3: Find p
p = distance from vertex to focus = 5 - 3 = 2
(Or: p = k - directrix = 3 - 1 = 2)
Step 4: Write the equation
(x - 2)² = 4(2)(y - 3)
(x - 2)² = 8(y - 3)
Answer: (x - 2)² = 8(y - 3)
Example 16: Horizontal Parabola with Focus and Vertex
Problem: Write the equation of a parabola with vertex (-1, 4) and focus (3, 4).
Solution:
Step 1: Determine orientation
Both points have the same y-coordinate (4), so this is a horizontal parabola.
Form: (y - k)² = 4p(x - h)
Step 2: Identify h and k
Vertex = (h, k) = (-1, 4), so h = -1, k = 4
Step 3: Find p
Focus = (h + p, k) = (3, 4)
h + p = 3
-1 + p = 3
p = 4
Step 4: Write the equation
(y - 4)² = 4(4)(x - (-1))
(y - 4)² = 16(x + 1)
Answer: (y - 4)² = 16(x + 1)
Example 17: Parabola with Given Width
Problem: A vertical parabola has vertex at (0, 0) and passes through the point (4, 2). Find its equation and the location of the focus.
Solution:
Step 1: Use the standard form for vertex at origin
x² = 4py
Step 2: Substitute the point (4, 2)
4² = 4p(2)
16 = 8p
p = 2
Step 3: Write the equation
x² = 4(2)y
x² = 8y
Step 4: Find the focus
Focus = (h, k + p) = (0, 0 + 2) = (0, 2)
Answer: Equation: x² = 8y; Focus: (0, 2)
Applications of Parabolas
Parabolas appear frequently in real-world applications due to their reflective property: any ray parallel to the axis of symmetry will reflect off the parabola and pass through the focus. This property is used in satellite dishes, telescopes, flashlights, and solar collectors.
Example 18: Satellite Dish Design
Problem: A satellite dish is shaped like a parabolic reflector. The dish is 12 feet across at its opening and 4 feet deep at its center. If we place the vertex at the origin with the parabola opening upward, find the equation of the parabola and determine where the receiver should be placed (at the focus).
Solution:
Step 1: Set up a coordinate system
Place the vertex at (0, 0) with the parabola opening upward.
The dish is 12 feet across, so it extends from x = -6 to x = 6.
At the edge (x = 6), the depth is y = 4.
Step 2: Use the standard form
x² = 4py
Substitute the point (6, 4):
6² = 4p(4)
36 = 16p
p = 36/16 = 9/4 = 2.25 feet
Step 3: Write the equation
x² = 4(2.25)y
x² = 9y
Step 4: Find the focus
Focus = (0, p) = (0, 2.25)
The receiver should be placed 2.25 feet above the vertex.
Answer: Equation: x² = 9y; Receiver location: 2.25 feet above the center
Example 19: Bridge Cable (Suspension Bridge)
Problem: The cables of a suspension bridge hang in the shape of a parabola. The towers supporting the cable are 400 feet apart and 100 feet tall. The lowest point of the cable is 10 feet above the roadway, which is at the midpoint between the towers. Find the equation of the parabola (with vertex at the lowest point) and determine the height of the cable at a point 50 feet from the center.
Solution:
Step 1: Set up coordinates
Place the vertex at (0, 10) where the cable is lowest.
The towers are 200 feet on each side of center, so at x = ±200, y = 100.
Step 2: Use the vertex form
(x - 0)² = 4p(y - 10)
x² = 4p(y - 10)
Step 3: Find p using the point (200, 100)
200² = 4p(100 - 10)
40,000 = 4p(90)
40,000 = 360p
p = 40,000/360 = 1000/9 ≈ 111.11 feet
Step 4: Write the equation
x² = 4(1000/9)(y - 10)
x² = (4000/9)(y - 10)
Step 5: Find height at x = 50
50² = (4000/9)(y - 10)
2500 = (4000/9)(y - 10)
2500 × 9/4000 = y - 10
22,500/4000 = y - 10
5.625 = y - 10
y = 15.625 feet
Answer: Equation: x² = (4000/9)(y - 10); Height at 50 ft from center: 15.625 feet
Example 20: Projectile Motion
Problem: A ball is thrown and follows a parabolic path. It reaches a maximum height of 20 feet when it is 30 feet horizontally from the starting point. If we place the vertex at the maximum height, find the equation and determine the horizontal distance when the ball returns to ground level (height = 0).
Solution:
Step 1: Set up coordinates
Place the vertex at (30, 20), the maximum height.
The parabola opens downward.
Step 2: Determine where the ball was thrown from
We need more information. Let's assume the ball was thrown from ground level (y = 0) at x = 0.
So we have the point (0, 0).
Step 3: Use the vertex form for a vertical parabola
(x - 30)² = 4p(y - 20)
Since it opens downward, p will be negative.
Step 4: Find p using the point (0, 0)
(0 - 30)² = 4p(0 - 20)
900 = 4p(-20)
900 = -80p
p = -900/80 = -11.25 feet
Step 5: Write the equation
(x - 30)² = 4(-11.25)(y - 20)
(x - 30)² = -45(y - 20)
Step 6: Find where the ball lands (y = 0)
(x - 30)² = -45(0 - 20)
(x - 30)² = -45(-20)
(x - 30)² = 900
x - 30 = ±30
x = 30 + 30 = 60 or x = 30 - 30 = 0
The ball lands at x = 60 feet (the other solution is where it was thrown).
Answer: Equation: (x - 30)² = -45(y - 20); Ball lands 60 feet from starting point
Example 21: Flashlight Reflector
Problem: A flashlight reflector is in the shape of a parabolic surface. The reflector is 6 cm in diameter and 3 cm deep. Where should the light bulb be positioned to ensure the light rays reflect parallel to the axis?
Solution:
Step 1: Set up coordinates
Place the vertex at the origin with the parabola opening to the right (or upward).
Let's use upward: vertex at (0, 0), parabola opens upward.
The reflector is 6 cm in diameter, so at the opening, x ranges from -3 to 3.
At x = 3, the depth is y = 3.
Step 2: Find p using the point (3, 3)
x² = 4py
3² = 4p(3)
9 = 12p
p = 9/12 = 3/4 = 0.75 cm
Step 3: Find the focus
Focus = (0, p) = (0, 0.75)
The light bulb should be positioned at the focus, 0.75 cm from the vertex.
Answer: Position the bulb 0.75 cm (or 7.5 mm) from the vertex of the reflector
Check Your Understanding
1. Find the vertex, focus, and directrix of the parabola (x + 3)² = 12(y - 1).
Answer: Vertex: (-3, 1); Focus: (-3, 4); Directrix: y = -2
This is a vertical parabola with h = -3, k = 1, and 4p = 12 so p = 3.
Focus = (h, k + p) = (-3, 1 + 3) = (-3, 4)
Directrix: y = k - p = 1 - 3 = -2
2. Which direction does the parabola (y - 2)² = -8(x + 1) open?
Answer: Opens to the left
This is a horizontal parabola with 4p = -8, so p = -2.
Since p < 0, the parabola opens to the left.
3. Write the equation of a parabola with vertex (4, -3) and focus (4, 1).
Answer: (x - 4)² = 16(y + 3)
Both points have x = 4, so this is vertical: (x - h)² = 4p(y - k)
h = 4, k = -3
Focus = (h, k + p) = (4, 1), so k + p = 1, giving -3 + p = 1, thus p = 4
Equation: (x - 4)² = 4(4)(y - (-3)) = 16(y + 3)
4. Convert y² - 6y - 4x + 5 = 0 to standard form and find the vertex.
Answer: (y - 3)² = 4(x + 1); Vertex: (-1, 3)
y² - 6y + 5 = 4x
Complete the square: (y² - 6y + 9) + 5 - 9 = 4x
(y - 3)² - 4 = 4x
(y - 3)² = 4x + 4 = 4(x + 1)
Vertex: (h, k) = (-1, 3)
5. A parabola has vertex at origin and passes through (3, -6). If it's a vertical parabola, what is its equation?
Answer: x² = -3y/2 or equivalently x² = -1.5y
Form: x² = 4py
Substitute (3, -6): 9 = 4p(-6)
9 = -24p
p = -9/24 = -3/8
Equation: x² = 4(-3/8)y = -3y/2
6. Find the directrix of the parabola y² = 16x.
Answer: x = -4
This is a horizontal parabola: (y - 0)² = 16(x - 0)
h = 0, k = 0, 4p = 16, so p = 4
Directrix: x = h - p = 0 - 4 = -4
7. What is the axis of symmetry for the parabola (x - 7)² = -20(y + 2)?
Answer: x = 7
This is a vertical parabola with vertex (7, -2).
The axis of symmetry is the vertical line through the vertex: x = 7
8. A parabola has focus at (5, 3) and directrix y = -1. Find its equation.
Answer: (x - 5)² = 8(y - 1)
Vertex is midway: k = (3 + (-1))/2 = 1, h = 5
Vertex: (5, 1)
p = k - directrix = 1 - (-1) = 2
Equation: (x - 5)² = 4(2)(y - 1) = 8(y - 1)
9. Convert x² + 10x + 2y + 21 = 0 to standard form.
Answer: (x + 5)² = -2(y - 2)
x² + 10x + 21 = -2y
Complete the square: (x + 5)² - 25 + 21 = -2y
(x + 5)² - 4 = -2y
(x + 5)² = -2y + 4 = -2(y - 2)
10. A satellite dish is 8 feet across and 2 feet deep. If the vertex is at the origin, where should the receiver be placed?
Answer: 2 feet above the vertex
The dish extends to (±4, 2). Using x² = 4py:
16 = 4p(2)
p = 2
Focus (receiver location) is at (0, 2), which is 2 feet above the vertex.
Key Takeaways
- A parabola is the set of all points equidistant from a focus (point) and directrix (line)
- The vertex is the point on the parabola closest to the directrix
- The parameter p is the directed distance from vertex to focus (and from vertex to directrix)
- Vertical parabolas: (x - h)² = 4p(y - k), with focus at (h, k + p) and directrix y = k - p
- Horizontal parabolas: (y - k)² = 4p(x - h), with focus at (h + p, k) and directrix x = h - p
- The sign of p determines direction: positive means up/right, negative means down/left
- The squared variable tells you orientation: x² means vertical, y² means horizontal
- Complete the square to convert from general form to standard form
- The focus-directrix property makes parabolas ideal for reflectors and collectors
- Always identify whether a parabola is vertical or horizontal before finding features