Parabola Anatomy Note Sheet Geometry & Algebra II • Intro Lesson
Parabola Anatomy: Focus & Directrix
Vertical parabolas, geometric definitions, and solving formulas
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The Geometric Definition
A parabola is the set of all points equidistant from a fixed point called the Focus and a fixed line called the Directrix: \( \text{Distance}(P, F) = \text{Distance}(P, d) \) .
Vertical Parabola Diagram \(d_1 = d_2\)
Axis: \(x = 0\) Directrix \(y = -p\) Focus \(F(0, p)\) Vertex \(V(0,0)\) Point \(P(x,y)\) \(d_1\) \(d_2\) \(p\) \(p\)
The vertex is always the exact midpoint between Focus and Directrix.
The Focus \((F)\) Point
A fixed interior point located at \((h, k + p)\) . For vertex \((0,0)\), it is \((0, p)\).
The Directrix \((d)\) Line
A horizontal line behind the vertex with equation \(y = -p\) (or \(y = k - p\)).
The Value of \(p\) Distance & Sign
\(|p|\) is the distance from Vertex to Focus. If \(p > 0\), opens UP ; if \(p < 0\), opens DOWN .
Finding the Vertex Midpoint
The Vertex is the midpoint between Focus and Directrix: \(y_{\text{vertex}} = \frac{y_{\text{focus}} + y_{\text{directrix}}}{2}\).
Formulas to Solve (Vertical Parabolas)
Vertex at \((0,0)\)
Quadratic Form \(y = \frac{1}{4p}x^2\) or \(y = ax^2\)
Since \(a = \frac{1}{4p}\), rearrange to solve directly for \(p\):
\(p = \frac{1}{4a}\) & \(a = \frac{1}{4p}\)
Example: \(y = \frac{1}{8}x^2 \implies 4p = 8 \implies p = 2\).
Conic Standard Form \(x^2 = 4py\)
The coefficient of \(y\) equals \(4p\). Divide it by \(4\):
\(p = \frac{\text{coefficient of } y}{4}\)
Example: \(x^2 = 12y \implies 4p = 12 \implies p = 3\).
1. Direction \(p > 0\) Up ↑ | \(p < 0\) Down ↓
2. Focus \(F(0, p)\)
3. Directrix \(y = -p\)
4. Vertex from \(F\) & \(d\) \(y_V = \frac{y_F + y_d}{2}\)
The Reflective Property (Real-World Application)
Headlights & Satellites
Incoming signals parallel to the axis bounce directly into the Focus . Conversely, a bulb placed at the focus emits parallel light beams.
Practice: Use the Formulas to Solve Work through each scenario
1. \(x^2 = 16y\) Equation \(\to\) Info
\(4p = 16 \implies p =\):
Opens:
Focus \((0, p)\): \((\) , \()\)
Directrix line: \(y =\)
2. \(y = \frac{1}{8}x^2\) Equation \(\to\) Info
\(4p = 8 \implies p =\):
Opens:
Focus \((0, p)\): \((\) , \()\)
Directrix line: \(y =\)
3. Focus \(F(0, 3)\), Directrix \(y = -3\) Find Vertex!
Midpoint \(y_V = \frac{3 + (-3)}{2}\): \(y_V =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal distance \(p\): \(p =\)
Equation \(y = \frac{1}{4p}x^2\): \(y =\)
Parabola Notes Answer Key Teacher Edition • Answer Key
Parabola Anatomy: Focus & Directrix
Complete solutions, teacher notes, and filled reference guide
Class Reference Geometry / Algebra II
The Geometric Definition
A parabola is the set of all points equidistant from a fixed point called the Focus and a fixed line called the Directrix: \( \text{Distance}(P, F) = \text{Distance}(P, d) \) .
Vertical Parabola Diagram KEY: \(d_1 = d_2 = 60\text{ px}\)
Axis: \(x = 0\) Directrix \(y = -p\) Focus \(F(0, p)\) Vertex \(V(0,0)\) Point \(P(x,y)\) \(d_1\) \(d_2\) \(p\) \(p\)
The vertex is always the exact midpoint between Focus and Directrix.
The Focus \((F)\) Point
A fixed interior point located at \((h, k + p)\) . For vertex \((0,0)\), it is \((0, p)\).
The Directrix \((d)\) Line
A horizontal line behind the vertex with equation \(y = -p\) (or \(y = k - p\)).
The Value of \(p\) Distance & Sign
\(|p|\) is the distance from Vertex to Focus. If \(p > 0\), opens UP ; if \(p < 0\), opens DOWN .
Finding the Vertex Midpoint
The Vertex is the midpoint between Focus and Directrix: \(y_{\text{vertex}} = \frac{y_{\text{focus}} + y_{\text{directrix}}}{2}\).
Formulas to Solve (Vertical Parabolas)
Vertex at \((0,0)\)
Quadratic Form \(y = \frac{1}{4p}x^2\) or \(y = ax^2\)
Since \(a = \frac{1}{4p}\), rearrange to solve directly for \(p\):
\(p = \frac{1}{4a}\) & \(a = \frac{1}{4p}\)
Teacher Note: Connects directly to Algebra 1 quadratic transformations!
Conic Standard Form \(x^2 = 4py\)
The coefficient of \(y\) equals \(4p\). Divide it by \(4\):
\(p = \frac{\text{coefficient of } y}{4}\)
Teacher Note: Remind students that if \(x^2 = -12y\), then \(p = -3\).
1. Direction \(p > 0\) Up ↑ | \(p < 0\) Down ↓
2. Focus \(F(0, p)\)
3. Directrix \(y = -p\)
4. Vertex from \(F\) & \(d\) \(y_V = \frac{y_F + y_d}{2}\)
The Reflective Property (Real-World Application)
Headlights & Satellites
Incoming signals parallel to the axis bounce directly into the Focus . Conversely, a bulb placed at the focus emits parallel light beams.
Answer Key: Practice Solutions Step-by-Step Worked Answers
1. \(x^2 = 16y\) Solved
\(4p = 16 \implies p =\): \(4\)
Parabola Vertex Practice Worksheet Conic Sections Practice
Finding Vertex from Focus & Directrix
Apply the midpoint formula and focal parameter relationships
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The Midpoint Rule: Vertex \(V(h, k)\) is halfway between Focus \(F(h, y_F)\) and Directrix \(y = y_d\):
\(k = \frac{y_F + y_d}{2}\) \(p = y_F - k\)
1 Part 1: Basic Practice (Vertices at the Origin)
Find the midpoint, determine \(p\), and write the equation
Problem 1: Focus \((0, 4)\), Directrix \(y = -4\)
Midpoint \(k = \frac{4 + (-4)}{2}\): \(k =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal length \(p = 4 - k\): \(p =\)
Opens:
Equation \(y = \frac{1}{4p}x^2\): \(y =\)
Problem 2: Focus \((0, -2)\), Directrix \(y = 2\)
Midpoint \(k = \frac{-2 + 2}{2}\): \(k =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal length \(p = -2 - k\): \(p =\)
Opens:
Equation \(y = \frac{1}{4p}x^2\): \(y =\)
Problem 3: Focus \((0, 5)\), Directrix \(y = -5\)
Midpoint \(k = \frac{5 + (-5)}{2}\): \(k =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal length \(p = 5 - k\): \(p =\)
Opens:
Equation \(x^2 = 4py\): \(x^2 =\)
2 Part 2: Stepped Practice (Shifted Vertices)
Remember: \(h\) matches the Focus \(x\)-coordinate; \(k\) is the midpoint
Problem 4: Focus \((0, 7)\), Directrix \(y = 1\)
Midpoint \(k = \frac{7 + 1}{2}\): \(k =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal length \(p = 7 - k\): \(p =\)
Opens:
Equation \(y - k = \frac{1}{4p}(x-h)^2\):
Problem 5: Focus \((2, 1)\), Directrix \(y = 7\)
Midpoint \(k = \frac{1 + 7}{2}\): \(k =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal length \(p = 1 - k\): \(p =\)
Opens:
Equation \(y - k = \frac{1}{4p}(x-h)^2\):
Problem 6: Focus \((-3, 0)\), Directrix \(y = -6\)
Midpoint \(k = \frac{0 + (-6)}{2}\): \(k =\)
Vertex \(V(h, k)\): \((\) , \()\)
Focal length \(p = 0 - k\): \(p =\)
Opens:
Equation \(y - k = \frac{1}{4p}(x-h)^2\):
3 Part 3: Conceptual Check & Reverse Challenge
Reasoning Check:
Can a parabola ever touch or cross its directrix? Why or why not? Explain using the distance definition.
Reverse Challenge:
A parabola has a Vertex at \((0, 2)\) and a Focus at \((0, 5)\). Find the value of \(p\) and write the equation of its directrix.
Parabola Vertex Practice Answer Key Teacher Edition • Answer Key
Finding Vertex from Focus & Directrix
Step-by-step solutions, midpoint calculations, and reasoning answers
Worksheet Solutions Parabola Practice Key
The Midpoint Rule: Vertex \(V(h, k)\) is halfway between Focus \(F(h, y_F)\) and Directrix \(y = y_d\):
\(k = \frac{y_F + y_d}{2}\) \(p = y_F - k\)
1 Part 1: Basic Practice (Vertices at the Origin)
All vertices are at \((0,0)\)
Problem 1: Focus \((0, 4)\), Directrix \(y = -4\)
Midpoint \(k = \frac{4 + (-4)}{2}\): \(k = 0\)
Vertex \(V(h, k)\): \((0, 0)\)
Focal length \(p = 4 - 0\): \(p = 4\)
Opens: Up (\(p > 0\))
Equation: \(y = \frac{1}{16}x^2\)
Problem 2: Focus \((0, -2)\), Directrix \(y = 2\)
Midpoint \(k = \frac{-2 + 2}{2}\): \(k = 0\)
Vertex \(V(h, k)\): \((0, 0)\)
Focal length \(p = -2 - 0\): \(p = -2\)
Opens: Down (\(p < 0\))
Equation: \(y = -\frac{1}{8}x^2\)
Problem 3: Focus \((0, 5)\), Directrix \(y = -5\)
Midpoint \(k = \frac{5 + (-5)}{2}\): \(k = 0\)
Vertex \(V(h, k)\): \((0, 0)\)
Focal length \(p = 5 - 0\): \(p = 5\)
Opens: Up (\(p > 0\))
Equation: \(x^2 = 20y\)
2 Part 2: Stepped Practice (Shifted Vertices)
Full Standard Vertex Form Solutions
Problem 4: Focus \((0, 7)\), Directrix \(y = 1\)
Midpoint \(k = \frac{7 + 1}{2}\): \(k = 4\)
Vertex \(V(h, k)\): \((0, 4)\)
Focal length \(p = 7 - 4\): \(p = 3\)
Opens: Up (\(p > 0\))
Equation: \(y - 4 = \frac{1}{12}x^2\) or \(x^2 = 12(y - 4)\)
Problem 5: Focus \((2, 1)\), Directrix \(y = 7\)
Midpoint \(k = \frac{1 + 7}{2}\): \(k = 4\)
Vertex \(V(h, k)\): \((2, 4)\)
Focal length \(p = 1 - 4\): \(p = -3\)
Opens: Down (\(p < 0\))
Equation: \(y - 4 = -\frac{1}{12}(x - 2)^2\)
Problem 6: Focus \((-3, 0)\), Directrix \(y = -6\)
Midpoint \(k = \frac{0 + (-6)}{2}\): \(k = -3\)
Vertex \(V(h, k)\): \((-3, -3)\)
Focal length \(p = 0 - (-3)\): \(p = 3\)
Opens: Up (\(p > 0\))
Equation: \(y + 3 = \frac{1}{12}(x + 3)^2\)
3 Part 3: Conceptual Check & Reverse Challenge Solutions
Reasoning Check Solution:
Can a parabola ever touch or cross its directrix?
No. If a point on the parabola were on the directrix, its distance to the directrix would be \(0\). By definition, its distance to the focus would also have to be \(0\), meaning the focus would lie on the directrix, which collapses the parabola into a degenerate line.
