HSG.GPE.A.2: Deriving the Equation of a Parabola from a Focus and Directrix
In plain English: HSG.GPE.A.2 is the Common Core geometry standard that asks students to derive the equation of a parabola from its focus and directrix. Students set the distance from a point (x, y) to the focus equal to its distance to the directrix, square both sides and simplify to a form such as (x - h)² = 4p(y - k). It is usually taught in Geometry or Algebra II.
Derive the equation of a parabola given a focus and directrix.
Common Core State Standards for Mathematics · Domain: Expressing Geometric Properties with Equations (GPE) · Cluster: Translate between the geometric description and the equation for a conic section Also written as HSG-GPE.A.2 or G-GPE.2 · Official standard
A parabola is the set of all points in a plane that are the same distance from a fixed point, the focus, and a fixed line, the directrix. In this lesson students turn that sentence into an equation. For a point P(x, y), they write the distance to the focus with the distance formula, write the perpendicular distance to the directrix, set the two equal, square both sides and simplify.
Students derive parabolas with horizontal directrices (which open up or down) and vertical directrices (which open left or right), with the vertex at the origin and elsewhere. They learn to predict the vertex as the midpoint between the focus and the directrix, and they check each derived equation with a point whose two distances they can measure.
Learning Objectives
By the end of this lesson, students will be able to:
State the focus-directrix definition of a parabola and write the distance from a point to the focus and to the directrix
Derive the equation of a parabola from a given focus and a horizontal directrix by setting the two distances equal and simplifying
Derive the equation of a parabola from a given focus and a vertical directrix, and explain why it opens left or right
Predict the vertex and opening direction from the focus and directrix, and check a derived equation with a point on the parabola
Prior Knowledge Required
Students should already be comfortable with:
The distance formula as an application of the Pythagorean Theorem 8.G.B.8
The distance from a point to a horizontal or vertical line
Expanding binomials such as (y - 5)² = y² - 10y + 25
The circle equation derived from a distance condition HSG.GPE.A.1
Draw the point F(0, 3) and the horizontal line y = -3 on a coordinate grid. Students answer on graph paper.
Warm-Up Prompt
"Find a point on the y-axis that is the same distance from F(0, 3) as from the line y = -3. Then find two points on the line y = 3 that are the same distance from F as from the line. What do you think the set of all such points looks like?"
The origin works: it is 3 units from F and 3 units from the line. On y = 3, the points (6, 3) and (-6, 3) are 6 units from F horizontally and 6 units above the line. Plot the three points and ask students to sketch a curve through them. Name the point F the focus and the line the directrix, and tell students they will now find the equation of the whole curve.
Direct Instruction20 minutes
The derivation. Continue with the warm-up focus F(0, 3) and directrix y = -3. Take a general point P(x, y) on the parabola and follow the steps:
Distance to the focus: by the distance formula, PF = √((x - 0)² + (y - 3)²).
Distance to the directrix: the shortest distance from P to the line y = -3 is the vertical distance, |y - (-3)| = |y + 3|.
Set them equal: √(x² + (y - 3)²) = |y + 3|. This is the definition of the parabola written in symbols.
Square both sides: x² + (y - 3)² = (y + 3)². Both sides were nonnegative, so no solutions are gained or lost.
Simplify: x² + y² - 6y + 9 = y² + 6y + 9, so x² = 12y, or y = x²/12.
Check with the warm-up points: (6, 3) gives 36 = 12(3). Point out that the y² terms cancel. That is why the result is a parabola and not a circle: only one variable stays squared. Then generalize with the class: for a focus (h, k + p) and directrix y = k - p, the same steps give (x - h)² = 4p(y - k), with vertex (h, k) halfway between the focus and the directrix. Work the examples below, and use Diagrams 1 and 2 for the second and fourth.
Vertex at the origin
Focus (0, 3), directrix y = -3.
Equation: x² + (y - 3)² = (y + 3)², so x² = 12y
Vertex not at the origin
Focus (2, 5), directrix y = 1. The vertex is halfway, at (2, 3).
After the third example, ask why the coefficient is negative: the focus is below the directrix, so the parabola opens downward. After the fourth, ask which variable is squared now and why (the directrix is vertical, so the x² terms cancel and y stays squared). The graph opens to the right and is not the graph of a function of x.
Guided Practice15 minutes
Pairs derive three equations on whiteboards. For each one, they first predict the vertex and the opening direction, then derive, then check a point.
Guided practice: predictions and derived equations
Focus and directrix
Vertex and opening
Derived equation
Focus (0, -1), directrix y = 1
(0, 0), opens down
x² = -4y
Focus (3, 4), directrix y = 0
(3, 2), opens up
(x - 3)² = 8(y - 2)
Focus (-2, 1), directrix x = 4
(1, 1), opens left
(y - 1)² = -12(x - 1)
Listen for these errors: measuring the distance to the directrix with the distance formula to some point on the line instead of perpendicular distance, expanding (y + 1)² as y² + 1, dropping a sign when moving terms, and using the vertical distance |y - k| when the directrix is a vertical line x = h.
Independent Practice15 minutes
Students work alone. Each derivation must show the distance equation before squaring.
Use Problem 1 to check that (10, 5) is on the parabola by computing both distances. (Distance to the focus is 10; distance to y = -5 is 10.)
Closure5 minutes
Exit ticket: (1) Derive the equation of the parabola with focus (1, 1) and directrix y = -1. (Answer: (x - 1)² = 4y.) (2) In one sentence, explain why the vertex of any parabola is halfway between the focus and the directrix. (The vertex is on the parabola, so its distances to the focus and to the directrix are equal, and it lies on the perpendicular from the focus to the directrix.)
Differentiation Strategies
For Struggling Students
Give a three-row organizer: distance to focus, distance to directrix, equation after squaring, so every derivation follows the same layout
Start with directrices that are the x-axis or y-axis so the distance to the directrix is simply |y| or |x|
Let students plot the focus, directrix and two or three equidistant points on graph paper before they derive the equation
For Advanced Students
Derive the general equation for focus (h, k + p) and directrix y = k - p, and explain what the sign of p tells you
Derive the parabola with focus (0, 0) and the slanted directrix x + y = 4, using the point-to-line distance |x + y - 4|/√2, and explain why the equation now has an xy-term
Show that the segment through the focus parallel to the directrix, with endpoints on the parabola, always has length 4|p|
Assessment Guidance
What to Look For
Strong derivations start from the definition: a written distance to the focus, a written distance to the directrix and an equals sign between them. Watch for students who jump straight to a memorized form such as x² = 4py; ask them to show where it comes from. Check that students use the perpendicular distance to the directrix (|y - k| for a horizontal line, |x - h| for a vertical one) and that they expand squared binomials completely. Ask every student to check the final equation with one point whose distances they compute directly.
02
Classroom Activities
3 Activities
1
Fold a Parabola
20 minPairs
Students fold wax paper so that a line lands on a point many times. The creases outline a parabola, which makes the focus-directrix definition physical before students derive its equation.
Procedure
On a sheet of wax paper about 30 cm long, draw a straight line 3 cm from the long bottom edge (the directrix) and mark a point 4 cm above the middle of that line (the focus)
Fold the paper so that one point of the line lands exactly on the focus, and crease. Repeat 20-30 times with different points along the line
The creases outline a curve. Trace it with a marker
Set up coordinates in centimeters with the origin halfway between the focus and the line, so the focus is (0, 2) and the directrix is y = -2. Derive the equation: x² + (y - 2)² = (y + 2)², so x² = 8y
Measure to check two points on the traced curve: (4, 2) should be 4 cm from the focus and 4 cm from the line, and (8, 8) should be 10 cm from both
Discussion Questions
When you fold a point of the line onto the focus, the crease is the perpendicular bisector of the segment joining them. Why does that mean points on the crease are equidistant from the focus and that point of the line?
What happens to the shape if the focus is closer to the directrix?
Modification for Distance Learning
Students use a free graphing tool: plot the focus and directrix, then add a movable point with its distances to the focus and to the line displayed. They drag the point until both distances match, record five such points, and compare them with their derived equation.
2
Focus-Directrix Card Derivations
20 minGroups of 3-4
Each group receives 4 cards, each giving one focus and one directrix. Groups plot the points they can find without an equation, then derive the equation and confirm that the plotted points satisfy it.
Cards
Card A: focus (0, 4), directrix y = 0. Result: x² = 8(y - 2), through (4, 4) and (-4, 4)
Card B: focus (1, -3), directrix y = 3. Result: (x - 1)² = -12y, through (7, -3) and (-5, -3)
Card C: focus (2, 0), directrix x = -4. Result: y² = 12(x + 1), through (2, 6) and (2, -6)
Card D: focus (-1, 2), directrix y = 6. Result: (x + 1)² = -8(y - 4), through (3, 2) and (-5, 2)
Procedure
Before any algebra, the group marks the vertex (the midpoint between the focus and the directrix) and the two points level with the focus, which are as far from the focus as the focus is from the directrix
Each student derives one card's equation from the distance equation, showing every step
The group checks that the three marked points satisfy each equation, then sketches all four parabolas
Groups rotate cards once and check another student's derivation line by line
Challenge Variation
Give groups a blank card. One student chooses a focus and a directrix; the others derive the equation. Then reverse the task: from the derived equation, can the group recover the focus and directrix that were chosen?
3
Design a Reflector
20 minPairs
A parabolic dish or reflector puts its receiver or bulb at the focus. Pairs derive the cross-section equation from the focus and directrix, then use it to find how deep the reflector must be.
Task A: A satellite dish
A dish has its vertex at the origin, and the receiver sits at the focus (0, 45), in centimeters. The directrix is y = -45
Derive the cross-section: x² + (y - 45)² = (y + 45)², so x² = 180y
The dish is 120 cm across. Find its depth at the rim: 60² = 180y, so y = 20 cm
Task B: A headlight reflector
The bulb is at the focus (0, 2.5), in centimeters, with directrix y = -2.5
Derive the cross-section: x² = 10y
The reflector is 12 cm across. Find its depth: 36 = 10y, so y = 3.6 cm
Discussion Questions
If the receiver moved farther from the vertex while the dish stayed 120 cm wide, would the dish get deeper or shallower? Use the equation to explain.
Why must the directrix be perpendicular to the axis of the dish?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Equal Distances to the Focus and the Directrix
The parabola with focus F(2, 5) and directrix y = 1, drawn to scale. The point P(6, 5) is 4 units from the focus and 4 units from the directrix. Setting the two distances equal for a general point (x, y) and squaring gives (x - 2)² = 8(y - 3), with vertex (2, 3).
Diagram 2: A Vertical Directrix Gives a Sideways Parabola
The parabola with focus F(3, -1) and directrix x = -1, drawn to scale. For P(3, 3), the distance to the focus and the horizontal distance to the directrix are both 4. The derivation gives (y + 1)² = 8(x - 1), which opens to the right.
04
Homework Assignment
~30 min
HSG.GPE.A.2 Homework: Parabolas from a Focus and Directrix
Directions: For every problem, write the distance from P(x, y) to the focus and to the directrix, set them equal, and show each step as you square and simplify. Name the vertex and the direction the parabola opens, and check your equation with one point.
Part 1: Deriving the Equation (Problems 1-3)
Derive the equation of the parabola with focus (0, -4) and directrix y = 4.
Derive the equation of the parabola with focus (-3, 6) and directrix y = 2. Write it in the form (x - h)² = 4p(y - k) and name the vertex.
Derive the equation of the parabola with focus (5, -2) and directrix x = -1. Explain why the parabola opens to the right.
Part 2: Using the Definition (Problems 4-6)
A point (x, y) moves so that it is always the same distance from (2, 3) as from the line y = -1. Derive the equation of its path, then find the two points on the path with y = 3 and verify that each is equally far from the point and the line.
A flashlight reflector has a parabolic cross-section with its vertex at the origin. The bulb is at the focus, 1.5 cm above the vertex. Derive the equation of the cross-section from the focus and directrix, then find how deep the reflector is if its opening is 12 cm wide.
A student derives the parabola with focus (1, -3) and directrix y = 1 and gets (x - 1)² = 8(y + 1). Without redoing the algebra, explain why the answer cannot be right. Then derive the correct equation.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Distance Setup
Both distances written correctly, including perpendicular distance to the directrix
One distance written incorrectly
No distance equation
Algebra
Squared and simplified correctly to a standard form
One expansion or sign error
Several errors or no simplification
Vertex and Direction
Vertex and opening direction correct and explained
Vertex or direction correct, not both
Missing
Checking and Context
Point checks shown; context answers include units
Checks or units incomplete
No checks
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so the quiz can be taken again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which description defines a parabola?
Answer: A
A parabola is the set of points equidistant from the focus (a point) and the directrix (a line). Choice B describes a circle. Choice C describes the perpendicular bisector of a segment, and choice D describes an ellipse.
Question 2 of 20 · Multiple Choice
What is the equation of the parabola with focus (0, 6) and directrix y = -6?
Answer: C
x² + (y - 6)² = (y + 6)² gives x² = 24y. Check: (12, 6) is 12 from the focus and 12 from the directrix, and 144 = 24(6). Choice A uses 2p instead of 4p. Choice B has the wrong variable squared: a horizontal directrix gives a parabola that opens up or down.
Question 3 of 20 · Multiple Choice
A parabola has focus (-2, 9) and directrix y = 1. What is its vertex?
Answer: B
The vertex is halfway between the focus and the directrix along the vertical line x = -2: y = (9 + 1)/2 = 5, so the vertex is (-2, 5). Choice A is the focus, choice C is a point on the directrix, and choice D swaps the coordinates.
Question 4 of 20 · Multiple Choice
What is the distance from a point P(x, y) to the directrix y = -5?
Answer: D
The shortest distance to a horizontal line is vertical: |y - (-5)| = |y + 5|. Choice A has the sign wrong. Choice B would be the distance to the vertical line x = -5. Choice C is the distance from P to one particular point, not to the line.
Question 5 of 20 · Multiple Choice
What is the equation of the parabola with focus (0, -2) and directrix y = 2?
Answer: B
x² + (y + 2)² = (y - 2)² simplifies to x² = -8y. The focus is below the directrix, so the parabola opens down and the coefficient is negative. Choice A has the wrong sign. Choice D uses 2p = -4 in place of 4p = -8.
Question 6 of 20 · Multiple Choice
A parabola has focus (4, 1) and directrix y = -3. Which equation states the definition correctly after squaring both sides?
Answer: A
PF² = (x - 4)² + (y - 1)², and the distance to y = -3 is |y + 3|, whose square is (y + 3)². Choice B measures to a vertical line. Choice C flips every sign. Choice D squares y + 3 as y² + 9, dropping the middle term 6y.
Question 7 of 20 · Multiple Choice
What is the equation of the parabola with focus (-2, 3) and directrix y = -1?
Answer: D
(x + 2)² + (y - 3)² = (y + 1)² gives (x + 2)² = 8y - 8 = 8(y - 1). The vertex (-2, 1) is halfway between y = 3 and y = -1. Choice A has the wrong sign for h. Choice C puts the focus where the vertex belongs. Choice B takes 4p to be the distance 4 between the focus and the directrix; that distance is 2p, so p = 2 and 4p = 8.
Question 8 of 20 · Multiple Choice
What is the equation of the parabola with focus (2, 0) and directrix x = -2?
Answer: C
The distance to the vertical line x = -2 is |x + 2|, so (x - 2)² + y² = (x + 2)², which gives y² = 8x. Choice A treats the directrix as horizontal. Choice D opens the wrong way: the focus is to the right of the directrix, so the parabola opens right.
Question 9 of 20 · Multiple Choice
A parabola has focus (0, -5) and directrix y = 1. Which statement is true?
Answer: B
A parabola always opens toward its focus and away from its directrix. The focus is below the directrix, so it opens down. The vertex is halfway: y = (-5 + 1)/2 = -2. The derivation gives x² = -12(y + 2). Choice C uses the focus as the vertex.
Question 10 of 20 · Multiple Choice
The parabola with focus (0, 4) and directrix y = -4 has equation x² = 16y. Which point is on it?
Answer: C
For (8, 4): 8² = 64 = 16(4). Its distance to the focus is 8 and its distance to the directrix is 8. Choice A gives 16 ≠ 64. Choice D gives 64 ≠ 32, and choice B gives 256 ≠ 64.
Question 11 of 20 · Multiple Choice
In the equation (x - h)² = 4p(y - k), derived from a focus and a directrix, what does p represent?
Answer: D
The focus is (h, k + p) and the directrix is y = k - p, so p is the signed distance from the vertex to the focus: positive when the parabola opens up, negative when it opens down. Choice A is 2|p|, twice as large. Choice B is k + p, not p.
Question 12 of 20 · Multiple Choice
A parabola has vertex (0, 0) and directrix y = 7. Where is its focus?
Answer: A
The vertex is halfway between the focus and the directrix. The directrix is 7 units above the vertex, so the focus is 7 units below it, at (0, -7), and the parabola opens down. Choice B is on the directrix. Choice D places the focus 14 units from the vertex, which would be the distance between the focus and the directrix, not from the vertex.
Question 13 of 20 · Multiple Choice
A student deriving a parabola writes √((x - 1)² + (y - 2)²) = |y + 4|, squares both sides and gets (x - 1)² + (y - 2)² = y² + 16. What is the error?
Answer: C
Squaring a binomial produces a middle term: (y + 4)² = y² + 8y + 16. Squaring both sides is allowed because both sides are nonnegative, so choice A is wrong. Choice D would be correct only for a vertical directrix x = -4.
Question 14 of 20 · Multiple Choice
What is the equation of the parabola with focus (1, 5) and directrix y = 3?
Answer: B
(x - 1)² + (y - 5)² = (y - 3)² gives (x - 1)² = 4y - 16 = 4(y - 4), with vertex (1, 4) and p = 1. Choice D has the wrong sign: the focus is above the directrix, so the parabola opens up. Choice C uses the focus as the vertex. Choice A takes 4p to be the distance 2 between the focus and the directrix, which is 2p.
Question 15 of 20 · Short Answer
Derive the equation of the parabola with focus (0, -3) and directrix y = 3. Show the distance equation before squaring.
√(x² + (y + 3)²) = |y - 3|. Squaring: x² + y² + 6y + 9 = y² - 6y + 9, so x² = -12y. It opens down with vertex (0, 0).
Question 16 of 20 · Short Answer
Derive the equation of the parabola with focus (2, -1) and directrix y = -5. Give the vertex.
(x - 2)² + (y + 1)² = (y + 5)² gives (x - 2)² = (y + 5)² - (y + 1)² = 8y + 24, so (x - 2)² = 8(y + 3). The vertex is (2, -3), halfway between y = -1 and y = -5.
Question 17 of 20 · Short Answer
Derive the equation of the parabola with focus (-4, 3) and directrix x = 2. Which way does it open?
The distance to x = 2 is |x - 2|. (x + 4)² + (y - 3)² = (x - 2)² gives (y - 3)² = (x - 2)² - (x + 4)² = -12x - 12, so (y - 3)² = -12(x + 1). The vertex is (-1, 3) and the parabola opens left, toward the focus.
Question 18 of 20 · Short Answer
When you derive a parabola's equation, you square both sides of √((x - h)² + (y - a)²) = |y - b|. Explain why this step does not add or lose any points.
Both sides are distances, so both are nonnegative. For nonnegative numbers, u = v exactly when u² = v². So the squared equation has exactly the same solutions as the distance equation, and the graph is exactly the set of points equidistant from the focus and the directrix.
Question 19 of 20 · Short Answer
A solar cooker has a parabolic cross-section with vertex at the origin. The cooking pot sits at the focus, 25 cm above the vertex. Derive the equation of the cross-section and find the cooker's depth if it is 80 cm across.
Focus (0, 25), directrix y = -25: x² + (y - 25)² = (y + 25)², so x² = 100y. At the rim, x = 40: 1600 = 100y, so the cooker is 16 cm deep.
Question 20 of 20 · Short Answer
A parabola has focus (0, p) and directrix y = -p and passes through the point (8, 2). Find p, and verify your answer with distances.
The derivation gives x² = 4py. Substituting (8, 2): 64 = 8p, so p = 8. Check: (8, 2) is √(64 + 36) = 10 units from the focus (0, 8) and 2 - (-8) = 10 units from the directrix y = -8.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GPE.A.2 mean?
HSG.GPE.A.2 means students can start from a focus and a directrix and derive the parabola's equation. They write the distance from a general point (x, y) to the focus, write its distance to the directrix, set the two equal, square and simplify. Memorizing a formula for the result is not enough; the standard asks for the derivation.
Is HSG.GPE.A.2 Geometry or Algebra 2?
It is a high school geometry standard and is usually taught in Geometry, right after the circle equation in HSG.GPE.A.1. Many Algebra II and Precalculus courses return to it when they study conic sections. Students need the distance formula and confidence expanding squared binomials.
What are the focus and the directrix of a parabola?
The focus is a fixed point and the directrix is a fixed line that does not pass through it. The parabola is every point that is exactly as far from the focus as from the directrix. The vertex is the point of the parabola closest to both, halfway between them.
How do you find the distance from a point to the directrix?
Use the perpendicular distance. For a horizontal directrix y = b, the distance from (x, y) is |y - b|. For a vertical directrix x = b, it is |x - b|. Using the distance formula to some other point on the line gives a longer, incorrect distance.
Why does the y² term cancel when the directrix is horizontal?
Both sides of the squared equation contain y²: the focus side from (y - a)², the directrix side from (y - b)². They cancel, leaving an equation in which only x is squared. That is why the result is a parabola rather than a circle. With a vertical directrix, the x² terms cancel instead and y stays squared.
What does p mean in (x - h)² = 4p(y - k)?
p is the directed distance from the vertex to the focus. The focus is (h, k + p) and the directrix is y = k - p, so the focus and directrix are 2|p| apart. When p is positive the parabola opens up; when p is negative it opens down.
Do students have to memorize the form (x - h)² = 4p(y - k)?
No. HSG.GPE.A.2 is about deriving the equation, so the distance equation is the starting point every time. The 4p form is a useful summary once students have derived it themselves, and it lets them check their work quickly.
What are common mistakes when deriving a parabola's equation?
Frequent errors include expanding (y + 3)² as y² + 9, measuring the distance to the directrix with the wrong variable, putting the focus at the vertex, and taking 4p to be the focus-to-directrix distance. A point check catches all of these: pick a point level with the focus and confirm both distances match.
Where are parabolas with a focus used in real life?
Satellite dishes, solar cookers, car headlights and flashlights use parabolic reflectors. Signals arriving parallel to the axis reflect to the focus, and light from a bulb at the focus reflects out in a parallel beam. Proving this reflection property goes beyond this standard, but finding where to put the receiver uses exactly the focus-directrix equation.
How does HSG.GPE.A.2 connect to other standards?
It uses the same distance method as the circle equation (HSG.GPE.A.1) and prepares students for ellipses and hyperbolas defined by two foci (HSG.GPE.A.3). It also gives a geometric meaning to the quadratic functions students graph in HSF.IF.C.7: every graph of y = ax² + bx + c is a parabola with a focus and a directrix.
07
Related Standards
6 standards
These standards connect to HSG.GPE.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.B.8Prerequisite
Use the Pythagorean Theorem to find the distance between two points on a grid
