HSG.GPE.A.3: Deriving the Equations of Ellipses and Hyperbolas from Their Foci
In plain English: HSG.GPE.A.3 is an advanced (+) Common Core geometry standard that asks students to derive the equations of ellipses and hyperbolas from their foci. Students write the condition that the sum (ellipse) or difference (hyperbola) of the distances to the two foci is constant, then square twice to reach x²/a² ± y²/b² = 1. It is usually taught in Precalculus.
(+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant.
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.3 or G-GPE.3 · Official standard
This is an advanced (+) standard, meant for students heading to Precalculus and beyond. An ellipse is the set of points whose distances to two fixed points, the foci, add up to a constant. A hyperbola is the set of points whose distances to the foci differ by a constant. Students turn each description into an equation: they write the two distances with the distance formula, set their sum (or difference) equal to 2a, isolate one square root, square, simplify, and square again.
The derivations end in x²/a² + y²/b² = 1 with b² = a² - c² for the ellipse and x²/a² - y²/b² = 1 with b² = c² - a² for the hyperbola, where the foci are (±c, 0). Students then apply the same process to foci on the y-axis and to foci that are not centered at the origin, and they explain with the triangle inequality why an ellipse needs a > c and a hyperbola needs a < c.
Learning Objectives
By the end of this lesson, students will be able to:
Write the focal-distance condition for an ellipse (constant sum 2a) and for a hyperbola (constant difference 2a) with the distance formula
Derive x²/a² + y²/b² = 1 with b² = a² - c² from foci (±c, 0) and a constant sum, by squaring twice
Derive x²/a² - y²/b² = 1 with b² = c² - a² from foci (±c, 0) and a constant difference
Write the equation of an ellipse or hyperbola whose foci are on the y-axis or centered at a point other than the origin
Use the triangle inequality to explain why the constant must be greater than the focal distance for an ellipse and less than it for a hyperbola
Prior Knowledge Required
Students should already be comfortable with:
The distance formula 8.G.B.8
Deriving the circle and parabola equations from a distance condition (HSG.GPE.A.1, HSG.GPE.A.2)
Solving equations with square roots by isolating and squaring HSA.REI.A.2
Plot F1(-8, 0) and F2(8, 0) on a grid. Students work in pairs.
Warm-Up Prompt
"Find every point on the x-axis whose distances to F1 and F2 add up to 20. Then find every point on the y-axis with the same property. Sketch a curve through your four points. What if the distances had to differ by 12 instead of adding up to 20?"
On the x-axis, (10, 0) and (-10, 0) work: 18 + 2 = 20. On the y-axis, (0, 6) and (0, -6) work, because each is √(64 + 36) = 10 from both foci. The sketch is an oval: an ellipse. For a difference of 12, the x-axis points are (6, 0) and (-6, 0), since 14 - 2 = 12, and no point on the y-axis works, since its two distances are always equal. Students will see that this second curve is a hyperbola with two separate branches.
Direct Instruction25 minutes
Part 1: Deriving the ellipse. Place the foci at F1(-c, 0) and F2(c, 0) and let the constant sum be 2a. For a point P(x, y) on the ellipse:
Name b: since a > c, the number a² - c² is positive; call it b². Divide by a²b² to get x²/a² + y²/b² = 1.
Use Diagram 1 to show what a, b and c mean: a is half the constant sum, c is the distance from the center to each focus, and at the top of the ellipse both focal distances equal a, so b² + c² = a². Part 2: Deriving the hyperbola. Repeat with the condition √((x + c)² + y²) - √((x - c)² + y²) = ±2a. The same two squarings give (c² - a²)x² - a²y² = a²(c² - a²). Now c > a, so c² - a² is positive; call it b² and get x²/a² - y²/b² = 1. Show Diagram 2. Then work the examples.
Ellipse, foci on the x-axis
Foci (±3, 0), sum of distances 10: a = 5, c = 3, b² = 25 - 9 = 16.
Equation: x²/25 + y²/16 = 1
Ellipse, foci on the y-axis
Foci (0, ±12), sum of distances 26: a = 13, c = 12, b² = 169 - 144 = 25. The major axis is vertical, so a² goes under y².
Equation: x²/25 + y²/169 = 1
Hyperbola, foci on the x-axis
Foci (±5, 0), difference of distances 6: a = 3, c = 5, b² = 25 - 9 = 16.
Equation: x²/9 - y²/16 = 1
Hyperbola, foci on the y-axis
Foci (0, ±10), difference of distances 12: a = 6, c = 10, b² = 100 - 36 = 64.
Equation: y²/36 - x²/64 = 1
Hyperbola, center not at the origin
Foci (-2, 1) and (8, 1), difference of distances 8. The center is the midpoint (3, 1), c = 5, a = 4, b² = 9.
Equation: (x - 3)²/16 - (y - 1)²/9 = 1
For the last example, explain that shifting both foci by the same amount shifts every point of the curve, so x and y are replaced by x - 3 and y - 1. Students can verify with the vertex (7, 1): its distances to the foci are 9 and 1, which differ by 8.
Guided Practice15 minutes
Pairs complete a table like the one below, then write out the full derivation for the first row.
Guided practice: from foci and constant to equation
Foci and constant
a, c, b²
Equation
Foci (±6, 0), sum 20
a = 10, c = 6, b² = 64
x²/100 + y²/64 = 1
Foci (0, ±17), difference 16
a = 8, c = 17, b² = 225
y²/64 - x²/225 = 1
Foci (0, ±2), sum 6
a = 3, c = 2, b² = 5
x²/5 + y²/9 = 1
Listen for these errors: using the constant itself as a instead of half of it, computing b² = a² - c² for a hyperbola, placing a² under x² when the foci are on the y-axis, and dropping the -4a√(...) middle term on the first squaring.
Independent Practice15 minutes
Students work alone and check each answer with a vertex, whose focal distances are easy to compute.
Ellipse with foci (±1, 0) and sum 4. (x²/4 + y²/3 = 1)
Hyperbola with foci (±6, 0) and difference 4. (x²/4 - y²/32 = 1)
Ellipse with foci (-1, 3) and (5, 3) and sum 10. (Center (2, 3), c = 3, a = 5, b = 4: (x - 2)²/25 + (y - 3)²/16 = 1)
Hyperbola with foci (2, -4) and (2, 6) and difference 6. (Center (2, 1), c = 5, a = 3, b = 4: (y - 1)²/9 - (x - 2)²/16 = 1)
Closure5-10 minutes
Exit ticket: (1) Two foci are 10 units apart. Explain why an ellipse with these foci must have a constant sum greater than 10, and a hyperbola must have a constant difference less than 10. (2) Write the hyperbola with foci (±15, 0) and difference 18. (Answer: a = 9, b² = 225 - 81 = 144, so x²/81 - y²/144 = 1.)
Differentiation Strategies
For Struggling Students
Give a fill-in derivation with the algebra partly done, so students focus on why each step is taken rather than on long expansions
Provide a summary card: ellipse, sum 2a, b² = a² - c²; hyperbola, difference 2a, b² = c² - a²; the larger denominator or the positive term tells the direction
Let students verify an equation numerically first: pick a vertex and a co-vertex, compute the focal distances and check the sum or difference
For Advanced Students
Show that squaring twice does not add extra points: every solution of x²/a² + y²/b² = 1 really has focal sum 2a
Derive the asymptotes y = ±(b/a)x of x²/a² - y²/b² = 1 by solving for y and letting x grow large
Show that the ellipse equation becomes the circle equation when the two foci coincide (c = 0)
Assessment Guidance
What to Look For
A complete derivation starts from the focal-distance condition, isolates one square root before squaring, and shows the second squaring. Students who jump to the final form have learned the answer, not the derivation the standard asks for. Check that students take a as half the constant, choose b² = a² - c² or b² = c² - a² correctly, and put a² under the variable along the axis through the foci. Ask each student to check an equation with one vertex: for an ellipse its focal distances add to 2a, and for a hyperbola they differ by 2a.
02
Classroom Activities
3 Activities
1
String-and-Pins Ellipse
20 minPairs
Students draw an ellipse with a string tied to two pins, which keeps the sum of the distances to the pins fixed. They then derive the equation for their drawing and compare its predicted height with a measurement.
Procedure
Push two pins into cardboard 8 cm apart. Tie the ends of a 10 cm piece of string to the pins
Pull the string tight with a pencil and trace all the way around. At every point, the distances to the pins add up to 10 cm
Set up coordinates with the origin midway between the pins: the foci are (±4, 0), 2a = 10 and c = 4. Derive the equation: x²/25 + y²/9 = 1
Predict the half-height of the ellipse (b = 3 cm) and measure it
Move the pins to 10 cm apart and use a 26 cm string. Derive the new equation (a = 13, c = 5, b = 12: x²/169 + y²/144 = 1) and check the half-height again
Discussion Questions
What happens to the shape as the pins move closer together with the same string? What if the pins were exactly 10 cm apart with the 10 cm string?
At the top of the ellipse, the string forms two equal halves. Why does that give b² + c² = a²?
Modification for Distance Learning
Students use a free graphing tool with two fixed points and a movable point that shows the sum of its distances to them. They drag the point to keep the sum at 10, trace the path, and compare it with the graph of x²/25 + y²/9 = 1.
2
Two Conics from Concentric Circles
20 minGroups of 3-4
Each student gets a sheet with two sets of concentric circles, radii 1 to 16 units, centered at two points 10 units apart. By marking intersections, groups draw an ellipse (constant sum) and a hyperbola (constant difference) and then derive both equations.
Procedure
In one color, mark every intersection where the radii add to 14, such as radius 4 around F1 and 10 around F2. Connect the marks: this is the ellipse with sum 14
In a second color, mark every intersection where the radii differ by 8, such as 12 and 4 or 5 and 13. Connect the marks: this is a hyperbola with two branches
Put the origin midway between the centers, so the foci are (±5, 0). Derive both equations: the ellipse has a = 7, b² = 49 - 25 = 24, so x²/49 + y²/24 = 1; the hyperbola has a = 4, b² = 25 - 16 = 9, so x²/16 - y²/9 = 1
Check each equation with its vertices: (±7, 0) and (±4, 0)
Discussion Questions
Why can no marks be made for a sum of 8 or a difference of 12 on this sheet?
Where do the two curves cross, and what are the two focal distances at a crossing point?
Challenge Variation
Groups choose their own sum and difference, predict whether a curve exists, mark it, and derive its equation. They then find the exact coordinates of one crossing point of their ellipse and hyperbola by solving the two equations together.
3
Locating Thunder and a Whispering Gallery
20 minPairs
Pairs derive a hyperbola from a time-difference measurement and an ellipse from a room design. Both tasks start from the foci and the constant sum or difference, as the standard requires.
Task A: Two listening stations
Two weather stations are 26 km apart, at (-13, 0) and (13, 0) on a map in kilometers. Thunder from one lightning strike reaches the east station about 29 seconds before the west station
Sound travels about 0.343 km per second, so the strike is about 29 × 0.343 ≈ 10 km farther from the west station than from the east one
Derive the hyperbola of possible locations: 2a = 10, c = 13, b² = 169 - 25 = 144, so x²/25 - y²/144 = 1, using only the branch closer to the east station (x > 0)
Task B: A whispering gallery
An elliptical room is 40 m long, and its foci are 12 m from the center. A whisper at one focus can be heard clearly at the other
Derive the floor outline: a = 20, c = 12, b² = 400 - 144 = 256, so x²/400 + y²/256 = 1
Find the width of the room: 2b = 32 m
Discussion Questions
Why does one time difference give a whole curve of possible locations rather than a single point? What extra information would pin down the strike?
In Task B, why is the path from one focus to the wall and on to the other focus always 40 m long?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: An Ellipse as a Constant Sum of Distances
The ellipse with foci (±3, 0) and constant sum 10, drawn to scale. The point P(4, 2.4) is 7.4 units from F1 and 2.6 units from F2, and 7.4 + 2.6 = 10. At the top point (0, 4), each focal distance is a = 5, which shows b² + c² = a².
Diagram 2: A Hyperbola as a Constant Difference of Distances
The hyperbola with foci (±5, 0) and constant difference 6, drawn to scale with its vertices (±3, 0). The point P(5, 16/3) is 34/3 units from F1 and 16/3 units from F2, a difference of 6. The dashed lines are the asymptotes, shown for reference.
04
Homework Assignment
~30 min
HSG.GPE.A.3 Homework: Ellipses and Hyperbolas from the Foci
Directions: Show the focal-distance condition for every problem. For Problems 1 and 4, show the full derivation with both squarings. For the other problems you may use the relationship between a, b and c that you derived, but state which one you use and why. Check each answer with a vertex.
Part 1: Ellipses (Problems 1-3)
Derive the equation of the ellipse with foci (-4, 0) and (4, 0) whose sum of distances from the foci is 12. Show both squaring steps.
Write the equation of the ellipse with foci (0, -8) and (0, 8) and sum of distances 20. Explain why a² goes under y².
Write the equation of the ellipse with foci (-3, -2) and (5, -2) and sum of distances 10. Name its center and check one vertex.
Part 2: Hyperbolas (Problems 4-6)
Derive the equation of the hyperbola with foci (-13, 0) and (13, 0) whose difference of distances from the foci is 24. Show both squaring steps.
Write the equation of the hyperbola with foci (0, -4) and (0, 4) and difference of distances 6.
Two listening posts are 20 km apart, at (-10, 0) and (10, 0) on a map in kilometers. A sound source is 8 km closer to the post at (10, 0) than to the other post. Write the equation of the hyperbola of possible locations and say which branch the source is on. Then explain why no hyperbola with these foci can have a difference of 20 km or more.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Focal Condition
Sum or difference of distances written correctly with the distance formula
Condition stated in words only or with one error
Missing
Derivation
Both squarings shown, algebra correct, b² defined
Derivation incomplete or one algebra error
No derivation
Equation
Correct a, b², orientation and center in every problem
One error in a, b², orientation or center
Several errors
Checking and Reasoning
Vertex checks shown; Problem 6 explained with the triangle inequality
Checks or explanation incomplete
No checks or explanation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Answer the questions in order. The score updates with each answer, and Reset quiz clears every answer for another attempt.
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
An ellipse is the set of points P for which the distances from P to the two foci have a constant:
Answer: A
For an ellipse, PF1 + PF2 is the same for every point, and that constant is 2a. Choice B describes a hyperbola. A constant product or ratio gives other curves that are not part of this standard.
Question 2 of 20 · Multiple Choice
A hyperbola is the set of points P for which:
Answer: D
For a hyperbola, the absolute difference of the focal distances is a constant 2a. The absolute value covers both branches: on one branch PF1 is larger, on the other PF2 is. Choice A is the ellipse, and choice C is the parabola from HSG.GPE.A.2.
Question 3 of 20 · Multiple Choice
What is the equation of the ellipse with foci (±12, 0) whose sum of distances from the foci is 30?
Answer: B
2a = 30 gives a = 15, and c = 12, so b² = a² - c² = 225 - 144 = 81. Choice A uses c² as b². Choice C adds a² and c², the hyperbola relationship. Choice D writes a and b instead of their squares.
Question 4 of 20 · Multiple Choice
What is the equation of the hyperbola with foci (±17, 0) whose difference of distances from the foci is 30?
Answer: C
2a = 30 gives a = 15, and c = 17, so b² = c² - a² = 289 - 225 = 64. Choice A is an ellipse equation. Choice B swaps a² and b², which would put the vertices at (±8, 0). Choice D uses b² = a² + c².
Question 5 of 20 · Multiple Choice
An ellipse has foci (±c, 0), constant sum 2a and co-vertices (0, ±b). If a = 17 and c = 15, what is b?
Answer: A
At the co-vertex (0, b), each focal distance is a, so b² + c² = a² and b² = 289 - 225 = 64, giving b = 8. Choice B subtracts a - c without squaring. Choice C adds a² and c², the hyperbola pattern. Choice D adds a and c.
Question 6 of 20 · Multiple Choice
A hyperbola x²/a² - y²/b² = 1 has foci (±c, 0), with a = 20 and b = 21. What is c?
Answer: C
The hyperbola derivation defines b² = c² - a², so c² = a² + b² = 400 + 441 = 841 and c = 29. Choice A subtracts a² from b², the ellipse pattern used backward. Choice B adds a and b instead of their squares, and choice D subtracts them.
Question 7 of 20 · Multiple Choice
Why must the constant sum for an ellipse be greater than the distance between its foci?
Answer: D
By the triangle inequality, PF1 + PF2 ≥ F1F2, with equality only for points on the segment between the foci. So 2a > 2c, and then b² = a² - c² is positive. Choice C has the inequality backward: for an ellipse a > c.
Question 8 of 20 · Multiple Choice
What is the equation of the ellipse with foci (0, ±8) whose sum of distances from the foci is 34?
Answer: B
a = 17, c = 8 and b² = 289 - 64 = 225. The foci are on the y-axis, so the major axis is vertical and a² = 289 goes under y². Choice A puts the major axis on the x-axis. Choice D is a hyperbola equation.
Question 9 of 20 · Multiple Choice
What is the equation of the hyperbola with foci (0, ±25) whose difference of distances from the foci is 14?
Answer: A
a = 7, c = 25 and b² = 625 - 49 = 576. The foci are on the y-axis, so the y²-term is the positive one. Choice B opens left and right instead of up and down. Choice C uses c² as b². Choice D is an ellipse equation.
Question 10 of 20 · Multiple Choice
An ellipse has equation x²/100 + y²/19 = 1. Where are its foci?
Answer: D
From the derivation, b² = a² - c², so c² = 100 - 19 = 81 and c = 9. The larger denominator is under x², so the foci are on the x-axis at (±9, 0). Choice A adds the denominators. Choice C gives the vertices, and choice B puts the foci on the wrong axis.
Question 11 of 20 · Multiple Choice
A hyperbola has foci F1(-7, 0) and F2(7, 0), and the difference of distances from the foci is 10. A point P on the hyperbola is 9 units from F1. How far is P from F2?
Answer: B
|PF1 - PF2| = 10 with PF1 = 9 means PF2 = 19 or PF2 = -1. A distance cannot be negative, so PF2 = 19, and P is on the branch nearer F1. Choice A comes from 10 - 9, but |9 - 1| = 8, not 10. Choice D adds 9 and the focal distance 14.
Question 12 of 20 · Multiple Choice
An ellipse has foci 12 units apart and a constant sum of distances of 18. A point P on the ellipse is 5 units from one focus. How far is P from the other focus?
Answer: C
The two distances add to 18, so the other distance is 18 - 5 = 13. The triangle inequality holds: 5 + 12 ≥ 13. Choice A adds instead of subtracting, choice D is half the sum (the value of a), and choice B subtracts 5 from 9.
Question 13 of 20 · Multiple Choice
In the hyperbola derivation with foci (±c, 0), you isolate √((x + c)² + y²) = 2a + √((x - c)² + y²) and square both sides. What do you get?
Answer: B
Squaring the binomial 2a + s, where s is the square root, gives 4a² + 4as + s². Choice A drops the middle term 4as, a common error when squaring a sum. Choice C has the sign of the middle term wrong; that sign belongs to the ellipse, where the root is subtracted. Choice D forgets to square 2a.
Question 14 of 20 · Multiple Choice
Why must a be less than c for a hyperbola with foci (±c, 0) and constant difference 2a?
Answer: D
The triangle inequality says the difference of two sides of a triangle is less than the third side, so 2a < 2c. That makes b² = c² - a² positive. Choice B is false in general: x²/9 - y²/4 = 1 has b < a. Choice C is false: the foci are inside the branches, not on them.
Question 15 of 20 · Short Answer
Derive the equation of the ellipse with foci (±2, 0) whose sum of distances from the foci is 8. Show the focal-distance condition and state a, c and b².
√((x + 2)² + y²) + √((x - 2)² + y²) = 8. Isolating and squaring twice gives (16 - 4)x² + 16y² = 16(16 - 4), that is 12x² + 16y² = 192. With a = 4, c = 2 and b² = 16 - 4 = 12: x²/16 + y²/12 = 1.
Question 16 of 20 · Short Answer
Derive the equation of the hyperbola with foci (±3, 0) whose difference of distances from the foci is 4.
√((x + 3)² + y²) - √((x - 3)² + y²) = ±4. Squaring twice gives (9 - 4)x² - 4y² = 4(9 - 4), that is 5x² - 4y² = 20. With a = 2, c = 3 and b² = 9 - 4 = 5: x²/4 - y²/5 = 1.
Question 17 of 20 · Short Answer
Write the equation of the ellipse with foci (2, 1) and (2, 9) whose sum of distances from the foci is 10.
The center is the midpoint (2, 5) and c = 4. With a = 5, b² = 25 - 16 = 9. The foci are on a vertical line, so a² goes under the y-term: (x - 2)²/9 + (y - 5)²/25 = 1. Check the vertex (2, 10): its distances to the foci are 9 and 1, which add to 10.
Question 18 of 20 · Short Answer
Write the equation of the hyperbola with foci (-4, 3) and (6, 3) whose difference of distances from the foci is 6.
The center is the midpoint (1, 3) and c = 5. With a = 3, b² = 25 - 9 = 16: (x - 1)²/9 - (y - 3)²/16 = 1. Check the vertex (4, 3): its distances to the foci are 8 and 2, which differ by 6.
Question 19 of 20 · Short Answer
The point (0, b) is on the ellipse with foci (±c, 0) and constant sum 2a. Use its two focal distances to explain why b² = a² - c². Then find b for foci (±20, 0) and sum 50, and write the equation.
(0, b) is the same distance from both foci, so each distance is half the sum, a. The right triangle with legs b and c and hypotenuse a gives b² + c² = a², so b² = a² - c². For foci (±20, 0) and sum 50: a = 25 and b² = 625 - 400 = 225, so b = 15 and x²/625 + y²/225 = 1.
Question 20 of 20 · Short Answer
Two foci are 10 units apart. Explain why no hyperbola with these foci can have a constant difference of distances equal to 12.
For any point P, the triangle inequality gives |PF1 - PF2| ≤ F1F2 = 10. A difference of 12 would need one side of triangle PF1F2 to exceed the sum of the other two, which is impossible. In symbols, 2a = 12 > 2c = 10 would make b² = c² - a² = 25 - 36 negative. No point has a difference of 12, so no such hyperbola exists.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GPE.A.3 mean?
HSG.GPE.A.3 means students can start from two foci and derive the equation of an ellipse or a hyperbola. For an ellipse they use the fact that the sum of the distances to the foci is constant; for a hyperbola, the difference. The derivation, with its two squarings, is the core of the standard.
What does the (+) in HSG.GPE.A.3 mean?
The (+) marks an advanced standard. Common Core describes these as additional mathematics that students should learn to take advanced courses such as calculus. Schools do not always require (+) standards for every student, so HSG.GPE.A.3 often appears in an honors or Precalculus course rather than in a standard Geometry course.
Is HSG.GPE.A.3 taught in Geometry or Precalculus?
It is usually taught in Precalculus, in a conic sections unit. Some honors Geometry or Algebra II courses include it after the circle (HSG.GPE.A.1) and the parabola (HSG.GPE.A.2).
Why do you have to square twice in the derivation?
The condition has two square roots. Isolating one and squaring removes it but leaves the other one behind, multiplied by 4a. After simplifying, you isolate that second root and square again. Trying to square with both roots on one side creates a messier equation with a cross term.
How do you know whether a² goes under x² or y²?
a² goes under the variable along the line through the foci. For foci (±c, 0), it goes under x²; for foci (0, ±c), under y². For an ellipse, a² is the larger denominator. For a hyperbola, a² is under the positive term, which may be smaller or larger than b².
Why is b² = a² - c² for an ellipse but c² = a² + b² for a hyperbola?
Both come from the same algebra, which produces the factor a² - c². For an ellipse a > c, so a² - c² is positive and is named b². For a hyperbola a < c, so c² - a² is the positive number named b². Geometrically, the ellipse has a right triangle with legs b and c and hypotenuse a, and for the hyperbola the rectangle with corners (±a, ±b) has half-diagonal c.
What are common mistakes when deriving these equations?
Frequent errors include using the whole constant as a instead of half of it, using the ellipse relationship for a hyperbola, dropping the middle term when squaring 2a - √(...), and putting a² under the wrong variable. A vertex check catches most of them.
Where are ellipses and hyperbolas used in real life?
Planets and many satellites move in elliptical orbits with the Sun or Earth at one focus. Elliptical rooms carry a whisper from one focus to the other. Hyperbolas describe all locations with the same difference in distance from two receivers, which older radio navigation systems and sound-ranging methods used to locate ships and sound sources.
Is a circle a special ellipse?
Yes. If the two foci are the same point, c = 0 and b² = a², so the ellipse equation becomes x²/a² + y²/a² = 1, or x² + y² = a². That is the circle equation from HSG.GPE.A.1, and the constant sum 2a is twice the radius.
How does HSG.GPE.A.3 connect to other standards?
It completes the conic sections cluster that begins with circles (HSG.GPE.A.1) and parabolas (HSG.GPE.A.2), using the same distance-formula method. The squaring steps rely on solving radical equations (HSA.REI.A.2), and the equations are used later in modeling and design problems (HSG.MG.A.3) and in calculus.
07
Related Standards
5 standards
These standards connect to HSG.GPE.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.GPE.A.1Prerequisite
Derive the equation of a circle and complete the square to find center and radius
