Defines the ellipse by a constant sum of distances to two foci, derives the standard equation from that definition, and identifies the centre, axes, vertices and foci from an equation. Covers both orientations, translated ellipses, and the relation among the three constants.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 10 — Analytic Geometry
§10.1 The Ellipse, pp. 1186-1207
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1186-1207 — the pages these objectives are drawn from
Warm-up
A circle has a familiar construction. An ellipse has one too.
Discussion prompt
With two pins, a loop of string and a pencil, what shape do you trace?
Hint: What stays constant as the pencil moves?
Answer:
Put the pins in, loop the string over both, and pull it taut with the pencil. Tracing round gives an ellipse.
What stays constant is the total string length, which is the sum of the distances from the pencil to the two pins.
So an ellipse is the set of points whose distances to two fixed points have a constant sum. That is the definition, and the whole section follows from it.
Concept
An ellipse is the set of points for which the distances to two fixed points, the foci, add to the same total. That total equals the length of the major axis.
ellipse — the set of all points whose distances to two fixed points have a constant sum
\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a^2=b^2+c^2 \]
Moving the two foci together turns the ellipse into a circle, since the constant sum becomes twice a single distance. A circle is the special case where the two foci coincide.
Figure (svg): An ellipse with two foci marked and a point on the curve joined to both, illustrating that the two distances have a constant sum
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1186-1191
Section
Section 1
Concept
Every point of the ellipse has the same total distance to the two foci, and that total is the major axis length. The definition is a statement about distances, not a formula.
Checking the definition at a vertex is what identifies the constant. From the near vertex the two distances are the semi-major axis minus and plus the focal distance, which add to twice the semi-major axis — the major axis length.
Figure (svg): An ellipse with two foci marked and a point on the curve joined to both, illustrating that the two distances have a constant sum
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1186-1192
Picture it
Two distances, one constant total.
Figure (svg): An ellipse with two foci marked and a point on the curve joined to both, illustrating that the two distances have a constant sum
Moving the point round the curve changes both distances but never their sum. That single constraint produces the whole shape.
Worked example
Check the definition at a vertex.
\[ \text{For an ellipse with semi-major axis } 5 \text{ and focal distance } 3, \text{ find the constant sum.} \]
Go to the near vertex
Why: On the major axis.
Find the near distance
Why: Semi-major minus focal.
\[ 5 - 3 = 2 \]
Find the far distance
Why: Semi-major plus focal.
\[ 5 + 3 = 8 \]
Add
Why: The constant sum.
\[ 10 \]
Figure (svg): An ellipse with two foci marked and a point on the curve joined to both, illustrating that the two distances have a constant sum
\[ \text{sum}=2a=10 \]
Verify: check at a co-vertex
Why: At the top of the ellipse both distances are equal, and each is the hypotenuse of a triangle with legs 4 and 3 — so each is 5 and the sum is 10, matching. Two different points giving the same sum confirms the constant.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1187-1189
Prediction
An ellipse has semi-major axis 7.
Predict first
What is the constant sum of distances?
Correct: Fourteen.
Why: The constant sum equals the major axis length, which is twice the semi-major axis. Checking at a vertex shows why: the two distances are the semi-major axis minus and plus the focal distance, and those add to twice the semi-major axis.
Worked example
The foci coincide.
\[ \text{What happens to the ellipse when the two foci are at the same point?} \]
Set the focal distance to zero
Why: Both foci at the centre.
\[ c = 0 \]
Apply the relation
Why: The semi-axes become equal.
\[ a = b \]
Read the condition
Why: The sum is twice one distance.
Identify the shape
Why: Points at a fixed distance.
Figure (svg): The solution to Worked example a circle as a special case shown as a ladder of expressions, one row per legal move
\[ x^2+y^2=a^2 \]
Verify: check the equation
Why: With the two denominators equal, the standard equation becomes the sum of the squares over a common value, which rearranges to the circle equation. So a circle is genuinely an ellipse rather than a separate shape.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1189-1192
Trap
\[ \text{an ellipse has a constant difference of distances to two points} \]
Recall the wrong operation
Why: Difference is used where sum belongs.
That describes a hyperbola, a completely different curve.
An ellipse has a constant sum. The string construction makes it visible: the loop's length is fixed and both segments are part of it.
A constant difference defines the hyperbola, which §10.2 develops.
The two definitions differ in one word and produce a closed curve and an unbounded one respectively, so the word carries a great deal.
Faded example
With semi-major axis 5 and focal distance 3.
Fill in the blanks
(5-3)+(5+3)=2+8=10
Why: At a vertex one focus is nearer and one further by exactly the focal distance, so the two terms cancel it and leave twice the semi-major axis. That is the constant for every point on the curve.
Sorting
One word separates two curves.
Sort into buckets
Sort each description.
Explain it to yourself
Two pins and a loop of string trace an ellipse.
Discussion prompt
Explain why that construction produces one.
Hint: What is fixed as the pencil moves?
Answer:
The string's length is fixed, and it consists of the segment from one pin to the pencil, the segment from the pencil to the other pin, and the fixed distance between the pins.
So the two variable segments have a constant sum — the string length minus the distance between the pins.
That is exactly the focal definition, with the pins as the foci. A good explanation notes that the construction is the definition made physical, which is why it produces the curve rather than merely approximating it.
Section
Section 2
Concept
In the standard form, each variable is squared over a denominator. The larger denominator sits under the variable running along the major axis, which fixes the orientation.
Fixing a as the larger is a convention, but it is universal and everything else depends on it. The relation among the three constants, the location of the foci, and the eccentricity all assume it.
Figure (svg): An ellipse with its centre, major and minor axes, vertices and co-vertices labelled
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1192-1197
Picture it
Centre, axes, vertices and co-vertices.
Figure (svg): An ellipse with its centre, major and minor axes, vertices and co-vertices labelled
The red line at the bottom is the convention that everything else rests on. Letting a be the smaller would invert the relation among the constants and put the foci in the wrong place.
Worked example
Compare the denominators.
\[ \text{Describe } \frac{x^2}{25}+\frac{y^2}{9}=1. \]
Compare the denominators
Why: Twenty-five is larger.
Read the orientation
Why: Major axis horizontal.
Find the semi-axes
Why: Square roots.
\[ a = 5, b = 3 \]
Locate the vertices
Why: Along the major axis.
\[ \text{at plus and minus } 5 \]
Figure (svg): An ellipse with its centre, major and minor axes, vertices and co-vertices labelled
\[ a=5,\;b=3,\;\text{horizontal} \]
Verify: check a point on the curve
Why: Substituting the vertex at 5 and 0 gives 25 over 25 plus 0, which is 1 — so it lies on the curve. The co-vertex at 0 and 3 gives 0 plus 9 over 9, also 1.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1193-1195
Prediction
The larger denominator sits under the vertical variable.
Predict first
Which way does the ellipse run?
Correct: Vertically, taller than it is wide.
Why: The larger denominator belongs to the variable running along the major axis. With it under the vertical variable, the ellipse extends further vertically than horizontally.
Worked example
The larger denominator has moved.
\[ \text{Describe } \frac{x^2}{16}+\frac{y^2}{49}=1. \]
Compare the denominators
Why: Forty-nine is larger.
Read the orientation
Why: Major axis vertical.
Find the semi-axes
Why: Square roots.
\[ a = 7, b = 4 \]
Locate the vertices
Why: Along the vertical axis.
\[ \text{at plus and minus } 7 \]
Figure (svg): The solution to Worked example the other orientation shown as a ladder of expressions, one row per legal move
\[ a=7,\;b=4,\;\text{vertical} \]
Verify: compare with the previous example
Why: The two equations differ only in which denominator is larger, and the ellipses differ only in orientation. That single comparison is the whole of the orientation question, which is why it is worth doing first.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1195-1197
Error analysis
A student reads an equation with the larger denominator second.
Annotate
On: \( \frac{x^2}{16}+\frac{y^2}{49}=1 \;\Longrightarrow\; a=4,\;b=7 \)
The letters carry meaning rather than position: a belongs to the major axis wherever that is. A negative value under a square root when finding the foci is the symptom that they were assigned by position instead.
Faded example
From an equation with denominators 25 and 9.
Fill in the blanks
a=\sqrt5=3, \quad b=\sqrt___=___
Why: The denominators are the squares of the semi-axes, so their square roots give the lengths. The larger root is a by convention, regardless of which term it came from.
Sorting
The larger denominator decides.
Sort into buckets
Sort each equation by its major axis direction.
Step zero
You are given an ellipse's equation to describe.
Discussion prompt
What do you compare first?
Hint: One comparison settles two things.
Answer:
Compare the two denominators. The larger one identifies both the orientation and which letter is a.
Everything after that depends on it: the vertices lie along the major axis, and the relation among the constants assumes a is the larger.
Assigning letters by position rather than by size is the error this prevents, and its symptom — a negative number under a square root — appears only later, when the foci are being located.
Section
Section 3
Concept
The three constants form a right triangle in which the semi-major axis is the hypotenuse, so its square equals the sum of the other two squares.
The relation looks like the Pythagorean theorem because it is one, applied to the triangle formed by the centre, a focus and a co-vertex. Seeing that triangle is what makes the formula memorable rather than a string of letters.
Figure (svg): A right triangle inside an ellipse showing the relation between the semi-major axis, semi-minor axis and focal distance
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1197-1201
Picture it
Semi-major axis as hypotenuse.
Figure (svg): A right triangle inside an ellipse showing the relation between the semi-major axis, semi-minor axis and focal distance
The triangle also shows why the semi-major axis exceeds both the others: a hypotenuse is always the longest side of a right triangle.
Worked example
Subtract, then take a root.
\[ \text{Find the foci of } \frac{x^2}{25}+\frac{y^2}{9}=1. \]
Identify the squares
Why: Larger first.
\[ a ^{2} = 25, b ^{2} = 9 \]
Apply the relation
Why: Subtract to isolate.
\[ c ^{2} = 25 - 9 = 16 \]
Take the root
Why: The focal distance.
\[ c = 4 \]
Place them
Why: On the major axis, either side.
\[ \text{at plus and minus } 4 \]
Figure (svg): A right triangle inside an ellipse showing the relation between the semi-major axis, semi-minor axis and focal distance
\[ (\pm 4,0) \]
Verify: check the constant sum
Why: From the focus at 4, the distances to the two vertices are 1 and 9, summing to 10 — which is twice the semi-major axis, as the definition requires. The foci are correctly placed.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1198-1200
Prediction
The three constants of an ellipse.
Predict first
Which is the largest?
Correct: The semi-major axis, being the hypotenuse.
Why: The relation puts the semi-major axis alone on one side as the hypotenuse of a right triangle, and a hypotenuse always exceeds either leg. Checking this ordering catches a reversed relation immediately.
Worked example
Run the relation the other way.
\[ \text{Find the equation with foci at } (\pm 3,0) \text{ and vertices at } (\pm 5,0). \]
Read the semi-major axis
Why: From the vertices.
\[ a = 5 \]
Read the focal distance
Why: From the foci.
\[ c = 3 \]
Apply the relation
Why: Solve for the other square.
\[ b ^{2} = 25 - 9 = 16 \]
Write the equation
Why: Larger denominator under x.
\[ x ^{2} / 25 + y ^{2} / 16 = 1 \]
Figure (svg): The solution to Worked example build an equation from the foci shown as a ladder of expressions, one row per legal move
\[ \frac{x^2}{25}+\frac{y^2}{16}=1 \]
Verify: check the foci come back
Why: Subtracting 16 from 25 gives 9, whose root is 3 — recovering the given focal distance. The relation works in both directions, which is what makes it usable for construction as well as description.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1200-1201
Trap
\[ c^2=a^2+b^2=25+9=34 \]
Add the two squares
Why: The relation is applied as though c were the hypotenuse.
The focal distance comes out larger than the semi-major axis, which puts the foci outside the ellipse.
The semi-major axis is the hypotenuse, so it is the one being squared alone on its side of the relation.
Finding the focal distance means subtracting: its square is the larger square minus the smaller.
Check that the focal distance is less than the semi-major axis. A focus outside the curve is impossible and the check catches the reversed relation immediately.
Faded example
With squares 25 and 9.
Fill in the blanks
c^2=25-9=16 \;\Longrightarrow\; c=4
Why: The focal distance is found by subtracting the smaller square from the larger, since the semi-major axis is the hypotenuse. Adding would place the foci outside the curve.
Sorting
The semi-major axis must be the largest.
Sort into buckets
Sort each set of values.
Explain it
The three constants satisfy a Pythagorean relation.
Discussion prompt
Explain to a classmate where the triangle is.
Hint: Which three points?
Answer:
Take the centre, one focus, and one co-vertex at the end of the minor axis. Those three points form a right triangle.
Its legs are the focal distance and the semi-minor axis, and its hypotenuse turns out to be the semi-major axis — which follows from the constant-sum definition applied at that co-vertex.
So the relation is the Pythagorean theorem for that triangle. A good explanation notes that seeing the triangle also explains the ordering: a hypotenuse is the longest side, so the semi-major axis exceeds both the others.
Section
Section 4
Concept
Moving an ellipse away from the origin replaces each variable by its difference from the centre's corresponding coordinate. Nothing else changes.
The sign convention inside the parentheses catches people out, exactly as it did in §1.5 with function transformations. A minus sign inside means the centre has moved in the positive direction.
Figure (svg): A contrast between the horizontal and vertical orientations of an ellipse, showing where the larger denominator sits
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1201-1205
Picture it
Translation preserves whichever of these applies.
Figure (svg): A contrast between the horizontal and vertical orientations of an ellipse, showing where the larger denominator sits
Moving the centre does not change which denominator is larger, so the orientation survives translation unchanged. Only the location moves.
Worked example
Read the centre from inside the parentheses.
\[ \text{Describe } \frac{(x-2)^2}{16}+\frac{(y+3)^2}{9}=1. \]
Read the centre
Why: Opposite signs to those shown.
\[ (2, -3) \]
Compare the denominators
Why: Sixteen is larger.
Find the semi-axes
Why: Square roots.
\[ a = 4, b = 3 \]
Locate the vertices
Why: Four either side of the centre.
\[ (-2, -3)\text{ and } (6, -3) \]
Figure (svg): A contrast between the horizontal and vertical orientations of an ellipse, showing where the larger denominator sits
\[ \text{centre }(2,-3),\; a=4,\; b=3 \]
Verify: check a vertex
Why: Substituting the point (6, -3) gives 16 over 16 plus zero, which is 1 — so it lies on the curve. The vertices are four units either side of the centre along the horizontal, as the semi-major axis requires.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1202-1204
Prediction
The equation shows x minus 5 and y plus 2.
Predict first
Where is the centre?
Correct: At five and negative two.
Why: The standard form subtracts the centre's coordinates, so the signs displayed are the opposite of the centre's. Substituting the centre should make both squared terms vanish, which is the check.
Worked example
Recover the standard form.
\[ \text{Put } 4x^2+9y^2-16x+18y-11=0 \text{ into standard form.} \]
Group and factor
Why: Coefficients out of each group.
\[ 4(x ^{2} - 4 x) + 9(y ^{2} + 2 y) = 11 \]
Complete both squares
Why: Half the coefficient, squared.
\[ 4(x - 2) ^{2} + 9(y + 1) ^{2} \]
Balance the constant
Why: Add what was really added.
\[ 11 + 16 + 9 = 36 \]
Divide through
Why: To make the right side one.
\[ \text{denominators } 9\text{ and } 4 \]
Figure (svg): The solution to Worked example complete the square shown as a ladder of expressions, one row per legal move
\[ \frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1 \]
Verify: check the centre satisfies neither term
Why: At the centre both squared terms vanish, giving zero rather than one — correct, since the centre is not on the ellipse. Substituting a vertex at (5, -1) gives 9 over 9, which is 1, confirming the form.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1204-1205
Error analysis
A student reads a translated ellipse's centre.
Annotate
On: \( \frac{(x-2)^2}{16}+\frac{(y+3)^2}{9}=1 \;\Longrightarrow\; \text{centre }(-2,3) \)
This is the same sign convention as every horizontal shift since §1.5. Substituting the claimed centre and checking that both squared terms vanish is a one-line test.
Faded example
For the expression x squared minus 4x.
Fill in the blanks
x^2-4x+4=(x-2)^2
Why: Half the coefficient of the linear term is 2, and its square is 4. Adding that inside the group means adding four times it to the equation, since the group carries a coefficient.
Sorting
Some features move and some do not.
Sort into buckets
Sort each feature.
Explain it to yourself
A minus inside means a positive shift.
Discussion prompt
Explain why.
Hint: What makes the squared term vanish?
Answer:
The squared term vanishes when the variable equals the centre's coordinate, and that happens when the expression inside is zero.
With x minus 2 inside, the term vanishes at x equal to 2 — so the centre is at 2, even though a minus sign is written.
It is the same convention as every horizontal shift since §1.5. A good explanation notes the check: substitute the claimed centre and both squared terms should come out zero.
Section
Section 5
Concept
The eccentricity is the ratio of the focal distance to the semi-major axis, and it measures how elongated the ellipse is. Zero is a circle and values near one are very stretched.
The eccentricity is dimensionless, so it describes the shape independently of size. Two ellipses with the same eccentricity are similar figures, one a scaled copy of the other.
Figure (svg): A right triangle inside an ellipse showing the relation between the semi-major axis, semi-minor axis and focal distance
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1205-1207
Picture it
Eccentricity is the ratio of two of them.
Figure (svg): A right triangle inside an ellipse showing the relation between the semi-major axis, semi-minor axis and focal distance
Because the focal distance is a leg and the semi-major axis the hypotenuse, their ratio is always less than one — which is why an ellipse's eccentricity is bounded.
Worked example
A ratio of two known lengths.
\[ \text{Find the eccentricity of } \frac{x^2}{25}+\frac{y^2}{9}=1. \]
Find the semi-major axis
Why: Root of the larger.
\[ 5 \]
Find the focal distance
Why: By the relation.
\[ 4 \]
Take the ratio
Why: Focal over semi-major.
\[ \frac{4}{5} \]
Interpret
Why: Fairly elongated.
\[ 0.8 \]
Figure (svg): A right triangle inside an ellipse showing the relation between the semi-major axis, semi-minor axis and focal distance
\[ e=\tfrac{4}{5}=0.8 \]
Verify: check the bound
Why: The eccentricity is less than one, as it must be since the focal distance is a leg of a right triangle whose hypotenuse is the semi-major axis. A value of 0.8 is close to one, matching an ellipse noticeably wider than it is tall.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1205-1206
Prediction
The two foci coincide.
Predict first
What is the eccentricity?
Correct: Zero.
Why: With the foci together the focal distance is zero, so the ratio is zero. That is the least elongated case, and values approaching one describe increasingly stretched ellipses.
Worked example
The primary body sits at a focus, not the centre.
\[ \text{A planet's orbit has semi-major axis } 10 \text{ and eccentricity } 0.2. \text{ Find its nearest and furthest distances.} \]
Find the focal distance
Why: Eccentricity times semi-major.
\[ 2 \]
Find the nearest distance
Why: Semi-major minus focal.
\[ 8 \]
Find the furthest
Why: Semi-major plus focal.
\[ 12 \]
Check the sum
Why: Twice the semi-major axis.
\[ 20 \]
Figure (svg): The solution to Worked example an orbit shown as a ladder of expressions, one row per legal move
\[ 8\text{ and }12 \]
Verify: check against the definition
Why: The two distances sum to 20, which is twice the semi-major axis — exactly the constant the focal definition requires. That both extremes lie on the major axis is why they are the nearest and furthest points.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1206-1207
Trap
\[ \text{the planet orbits the star, so the star is at the ellipse's centre} \]
Place the massive body at the geometric centre
Why: The centre is assumed to be the natural place for it.
The orbit's nearest and furthest distances then come out equal, which is a circle.
The primary body is at a focus, not the centre. That is Kepler's first law.
Which is why an orbit has a nearest and a furthest point at all — they differ by twice the focal distance.
The centre of an orbital ellipse is empty space. Placing the star there would make every orbit circular, contradicting observation.
Faded example
With focal distance 4 and semi-major axis 5.
Fill in the blanks
e=\frac50.8=\frac______}=___
Why: The eccentricity is the focal distance divided by the semi-major axis. It is always less than one for an ellipse, since the focal distance is a leg and the semi-major axis the hypotenuse.
Sorting
Zero is circular and near one is elongated.
Sort into buckets
Sort each value.
Explain it
An ellipse's eccentricity is always below one.
Discussion prompt
Explain to a classmate why.
Hint: Which sides of the triangle are involved?
Answer:
The eccentricity is the focal distance over the semi-major axis, and in the defining right triangle those are a leg and the hypotenuse.
A leg is always shorter than the hypotenuse, so the ratio is always less than one. It reaches zero only when the leg vanishes, which is the circle.
So the bound is geometric rather than a definition. A good explanation adds the contrast with the hyperbola, whose eccentricity exceeds one because its relation puts the focal distance in the hypotenuse's place.
Comparison
Fill the blanks from memory. Each constant has one meaning.
Comparison matrix
| a | b | c | |
|---|---|---|---|
| what it measures | the semi-major axis | the semi-minor axis | the focal distance |
| in the equation | root of the larger denominator | root of the smaller | not shown; computed |
| in the triangle | the hypotenuse | a leg | a leg |
| size ordering | the largest | less than a | less than a |
The third row explains the fourth. The semi-major axis is a hypotenuse, so it exceeds both of the others in every ellipse.
Pattern
Five steps, and the second decides everything after it.
Step 2 before step 4 matters: the relation assumes a is the larger, so assigning the letters by position gives a negative square root.
OpenStax Algebra and Trigonometry 2e, §12.1 The Ellipse §12.1
Check
The definition.
Check your understanding
What is constant for every point on an ellipse?
Answer: A
Why: The constant sum is the defining property, and it equals the major axis length. A constant difference would define a hyperbola, and a constant distance to a centre would give a circle.
Check
The relation.
Check your understanding
How is the focal distance found from the semi-axes?
Answer: A
Why: The semi-major axis is the hypotenuse of the defining right triangle, so isolating the focal distance means subtracting. Adding would give a focal distance exceeding the semi-major axis, putting the foci outside the curve.
Check
Orientation.
Check your understanding
The larger denominator sits under the vertical variable. What follows?
Answer: A
Why: The larger denominator belongs to the variable running along the major axis, so the ellipse extends further vertically. That single comparison settles the orientation and identifies which letter is a.
Real world
A whispering gallery works because of the focal definition.
Discussion prompt
In an elliptical room, a whisper at one focus is clearly audible at the other. Why?
Hint: What happens to a sound wave reflecting off the wall?
Answer:
Sound leaving one focus reflects off the wall and travels to the other focus, whatever direction it started in — a geometric property of the ellipse.
And every such path has the same total length, by the defining constant sum. So all the reflected waves arrive at the second focus at the same moment.
Arriving in step, they reinforce each other rather than smearing out, which is why the whisper is audible. The constant-sum definition is doing the work directly, and the same principle focuses shock waves in medical lithotripsy.
Commit first
State your confidence along with your answer.
Predict first
Why is the semi-major axis always the largest of the three constants?
Correct: It is the hypotenuse of the defining right triangle.
Why: The centre, a focus and a co-vertex form a right triangle whose legs are the focal distance and the semi-minor axis, with the semi-major axis as hypotenuse. A hypotenuse always exceeds either leg.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate assigned a to the first denominator and got a negative number under a square root. Explain what happened.
Hint: What does a mean?
Answer:
The letter a always means the larger semi-axis, not the first one written. Assigning by position rather than by size is the error.
With the letters swapped, subtracting the larger square from the smaller gives a negative value, and the square root fails.
So the negative under the root is the symptom, not the disease. A good explanation adds the fix: compare the denominators before assigning any letters, since that one comparison also settles the orientation.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the ellipse and hyperbola differ most, so getting it firm now pays off in the next section. The fourth is where most of the algebra lives.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw an ellipse with its centre, foci, vertices and co-vertices labelled, and mark the right triangle that gives the relation among the constants. Beside it, write the standard equation in both orientations and note which denominator is larger in each. Underneath, complete the square on one expanded equation to recover standard form.
If your triangle has the semi-major axis as its hypotenuse and your two orientations differ only in which denominator is larger, the section's two decisive facts are on the page.
Recap
Five things, and the second is the one to do first every time.
| if you remember one thing | it should be this |
|---|---|
| about the definition | constant sum, which equals the major axis length |
| about the letters | a is the larger, wherever it appears in the equation |
| about the relation | the semi-major axis is the hypotenuse, so subtract to find c |
| about orbits | the primary body sits at a focus, never at the centre |
Section 10.2 changes one word of the definition — sum becomes difference — and produces the hyperbola, an unbounded curve whose relation among the constants is rearranged accordingly.
OpenStax, Precalculus, §10.1 The Ellipse §10.1, pp. 1186-1207 — everything on these slides traces back here
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