Parabola Blueprint Slides Drafting Session 01
Parabola Blueprint
Building the equation from the ground up using the Focus and Directrix.
The Secret Definition
A parabola isn't just a "U" shape. It is the set of all points that are equidistant from:
The Focus (a fixed point)
The Directrix (a fixed line)
FIG. A: EQUIDISTANT PROPERTY
d₁ d₂
d₁ = d₂
The Essential Tool
To build the equation, we need the Distance Formula:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
Point 1
Focus (0, p)
Point 2
Directrix (x, -p)
Step-by-Step Build
01 / 04
1
Set the Distances Equal
Start with our "Secret Definition": Distance to Focus = Distance to Directrix.
\[\sqrt{(x-0)^2 + (y-p)^2} = \sqrt{(x-x)^2 + (y - (-p))^2}\]
Square Both Sides
02 / 04
2
Eliminate the Radicals
Square both sides and simplify the zeroes.
\[x^2 + (y-p)^2 = (y+p)^2\]
3
Expand the Squares
Remember: \((y-p)^2 = y^2 - 2py + p^2\)
\[x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2\]
The Result
04 / 04
Cancel shared terms:
\[x^2 + \cancel{y^2} - 2py + \cancel{p^2} = \cancel{y^2} + 2py + \cancel{p^2}\]
FINAL FORM
\[y = \frac{1}{4p}x^2\]
where p is the distance from vertex to focus.
On Your Blueprint
Find the equation of a parabola with:
Focus: (0, 3)
Directrix: y = -3
Key Hint:
If Focus is (0, 3), then p = 3
\[y = \frac{1}{4(\_\_)}x^2 \longrightarrow y = \frac{1}{12}x^2\]
Parabola Blueprint Teacher Guide Facilitation Guide
Parabola Blueprint Intervention
TIER 2 INTERVENTION
SESSION OBJECTIVE
Students will derive the equation of a parabola \((y = ax^2)\) using the distance formula and the locus definition (equidistant from focus and directrix).
COLORADO STANDARD
HS.G-GPE.A.2: Derive the equation of a parabola given a focus and directrix.
Small Group Facilitation Path
1
Visual Anchor (5 mins)
Use Slide 2 to introduce the equidistant property . Ask students: "If I move a point along this curve, what happens to its distance from the focus compared to the line?"
Key Question: "What does 'equidistant' mean in your own words?"
2
The Setup (10 mins)
On the worksheet, have students label the Focus \((0, p)\) and a point on the directrix \((x, -p)\). Help them see that the point on the directrix is always directly below the point \((x, y)\) on the parabola.
Watch for: Students confusing the directrix \(y = -p\) with a point. Remind them the distance to a line is the perpendicular (vertical) distance.
3
The Algebra Heavy Lift (15 mins)
Guide students through the derivation on Slides 4-6. This is where Tier 2 support is critical. Walk through the squaring of binomials slowly.
Prompt: "Look at all these terms. Most of them appear on both sides of the equal sign. What happens when we subtract them from both sides?"
Misconception Blueprint
The "p" Confusion
Students may think \(p\) is the focus point, rather than the distance from the vertex to the focus. Use the visual to show \(p\) is a length.
Squaring Binomials
Students often write \((y-p)^2\) as \(y^2 - p^2\). Remind them of FOIL or the box method: \(y^2 - 2py + p^2\).
Progress Monitoring Checklist
Definition Check
Student can explain that the distance from point P to the Focus and P to the Directrix are equal.
Formula Setup
Student can correctly plug focus coordinates and directrix height into the distance formula.
Application Success
Student can find the final equation for a simple focus like (0, 2) and directrix y = -2.
Next Steps for Differentiation
If struggling: Provide a worksheet with the algebra already expanded, focus only on the cancellation step.
If mastered: Ask students how the equation changes if the parabola opens horizontally (focus on x-axis).
Parabola Blueprint Worksheet Parabola Blueprint
ARCHITECTURAL DERIVATION WORKSHEET
NAME:
DATE:
1. The Vision
A parabola is built from a Focus and a Directrix . Every point \((x, y)\) on the parabola must be the same distance from both.
Distance to Focus \(d_1\)
=
Distance to Directrix \(d_2\)
2. The Construction
Follow the steps to build the formula. Fill in the missing blueprints.
Step A: Set up the Distance Formula
Focus at \((0, p)\). Directrix at \(y = -p\).
\[\sqrt{(x-0)^2 + (y-p)^2} = \sqrt{(x-x)^2 + (y - ( \quad ))^2}\]
Step B: Square both sides and simplify
\[x^2 + (y - p)^2 = (y + \quad )^2\]
Step C: Expand the binomials
\[x^2 + y^2 - 2py + p^2 = y^2 + \quad \quad + p^2\]
Step D: Cancel common terms and solve for \(y\)
3. Field Applications
JOB #1: FOCUS (0, 2) | DIRECTRIX y = -2
1. Identify \(p\):
\(p = \) ________
2. Plug into the final form \(y = \frac{1}{4p}x^2\):
\(y = \) ________________
Rough Sketch
JOB #2: FOCUS (0, 5) | DIRECTRIX y = -5
Calculate the equation:
Final Equation: ________________________
SITE REFLECTION
If the Focus point moves further away from the Directrix (like in Job #2 vs Job #1), does the parabola get wider or narrower? Explain.
Parabola Blueprint Answer Key Parabola Blueprint
ARCHITECTURAL DERIVATION - ANSWER KEY
Teacher Reference
2. The Construction - Solved
Step A: Set up the Distance Formula
\[\sqrt{(x-0)^2 + (y-p)^2} = \sqrt{(x-x)^2 + (y - ( \mathbf{-p} ))^2}\]
Step B: Square both sides and simplify
\[x^2 + (y - p)^2 = (y + \mathbf{p} )^2\]
Step C: Expand the binomials
\[x^2 + y^2 - 2py + p^2 = y^2 + \mathbf{2py} + p^2\]
Step D: Cancel common terms and solve for \(y\)
\[x^2 - 2py = 2py\]
\[y = \frac{1}{4p}x^2\]
3. Field Applications - Solved
JOB #1: FOCUS (0, 2) | DIRECTRIX y = -2
1. Identify \(p\):
\(p = 2\)
2. Final Form:
\(y = \frac{1}{8}x^2\)
Correct Sketch
JOB #2: FOCUS (0, 5) | DIRECTRIX y = -5
Calculate the equation:
p = 5
Final Equation: y = (1/20)x²
SITE REFLECTION ANSWER
"The parabola gets wider . As the distance \(p\) increases, the fraction \(\frac{1}{4p}\) gets smaller (closer to zero). A smaller 'a' value in \(y = ax^2\) creates a wider vertical compression."
