The two-foci definition of an ellipse and its vocabulary, the two standard equations with centre at the origin, identifying the major axis and locating vertices, co-vertices and foci, writing an equation from a vertex together with a co-vertex or a focus, and finding the area of an elliptical region.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Graph and Write Equations of Ellipses
Objectives
Five outcomes. Two foci instead of one centre, and a circle stretched along one direction.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-637 — the lesson these objectives are drawn from
Warm-up
Lesson 9.3 defined a circle by one fixed distance from one centre.
Discussion prompt
Pin a loop of string around two tacks, put a pencil in the loop and pull it taut. As the pencil moves, what stays constant, and what shape does it trace?
Hint: The string has a fixed length.
Answer:
The two distances from the pencil to the tacks always add to the same total: the string's length minus the gap between the tacks.
\[ d_1 + d_2 = \text{constant} \]
The curve is an ellipse, and the tacks are its foci. Bringing the two tacks together gives a circle, so a circle is the special case where the two foci coincide.
Concept
An ellipse is the set of points for which the distances to two fixed points, the foci, add to a constant. That constant is the length of the major axis, and the two axes give the two denominators of the standard equation.
ellipse — The set of all points in a plane for which the sum of the distances to two fixed points, called the foci, is constant. The vertices lie on the major axis and the co-vertices on the minor axis.
\[ \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1, \quad a > b > 0, \quad c^2 = a^2-b^2 \]
The right-hand side is 1 rather than a squared length, so both denominators can be read off directly — and the larger one always sits under the major axis's variable.
Figure (svg): Two columns comparing a circle with an ellipse
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-634
Section
Section 1
Concept
The line through the two foci meets the ellipse at the vertices, joined by the major axis whose midpoint is the centre. The perpendicular line through the centre meets it at the co-vertices, joined by the minor axis.
\[ d_1+d_2 = 2a \]
The major axis is always the longer one and always contains the foci. That pairing is what makes the orientation readable from a single glance at the equation.
Figure (svg): An ellipse with two foci and a point on it, showing the two distances that always sum to the same value
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-634 — Definition of an ellipse
Picture it
An ellipse with two of its points and both pairs of distances.
Figure (svg): An ellipse with two foci and a point on it, showing the two distances that always sum to the same value
At the top the two distances are 5 and 5; at the right-hand point they are 7 and 3. Both add to 10, which is the length of the major axis.
Worked example
Verifying the definition at two points of an ellipse with foci 4 units from the centre.
\[ \text{For foci } (\pm 4, 0) \text{ and } a = 5, \text{ check the points } (0,3) \text{ and } (5,0). \]
Top point: distance to each focus
Why: Four across and 3 up, so 16 plus 9.
\[ 5\text{ and } 5 \]
Top point: add them
Why: Five plus 5.
\[ 10 \]
Right vertex: distances
Why: From 5 comma 0 to each focus along the axis.
\[ 1\text{ and } 9 \]
Right vertex: add them
Why: One plus 9.
\[ 10 \]
Figure (svg): An ellipse with two foci and a point on it, showing the two distances that always sum to the same value
\[ d_1+d_2 = 10 = 2a \]
Verify: explain why the sum equals 2a
Why: At a vertex the two distances are a minus c and a plus c, and those add to 2a whatever c is. So the constant sum is always the major axis's full length — which is why a is recoverable from the definition without measuring anything else.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-634
Matching
For a horizontal ellipse centred at the origin.
Match the pairs
Why: Three of the four sit on the major axis, and only the co-vertices are on the minor one. That concentration is why identifying the major axis first settles most of the work.
Worked example
Reading the labelled diagram of the key concept.
\[ \text{For } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ with } a>b>0, \text{ name the centre, vertices, co-vertices and foci.} \]
The centre
Why: The midpoint of the major axis.
\[ (0, 0) \]
The vertices
Why: On the horizontal axis, a units out.
\[ (a, 0)\text{ and } (-a, 0) \]
The co-vertices
Why: On the vertical axis, b units out.
\[ (0, b)\text{ and } (0, -b) \]
The foci
Why: On the major axis, c units out, with c squared equal to a squared minus b squared.
\[ (c, 0)\text{ and } (-c, 0) \]
Figure (svg): The parts of an ellipse labelled, in both the horizontal and vertical orientations
\[ (\pm a, 0), \; (0, \pm b), \; (\pm c, 0) \]
Verify: check that c is smaller than a
Why: Since c squared equals a squared minus b squared and b is positive, c must be less than a — so the foci always lie strictly inside the ellipse, between the centre and the vertices. A computed c larger than a would signal that the roles of a and b had been swapped.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-634
Trap
\[ \frac{x^2}{25}+\frac{y^2}{4}=1 \]
Place the foci on the vertical axis
Why: The smaller denominator is taken to mark the foci.
\[ \text{foci } (0, \pm\sqrt{21}) \quad \text{(wrong)} \]
The square root of 21 is about 4.6, but the ellipse only reaches 2 in the vertical direction — the foci would lie outside the curve entirely.
\[ \text{foci } (\pm\sqrt{21}, 0) \]
The foci always lie on the MAJOR axis
Why: That is the axis with the larger denominator, and the only one long enough to contain them.
\[ \sqrt{21} \approx 4.6 < 5 = a \quad \checkmark \]
Checking that c is less than a catches this instantly, and it is the reason the definition puts the foci on the longer axis in the first place.
Fill the middle
At a vertex of an ellipse.
Fill in the blanks
(a-c)+(a+c) = 2a
Why: The c terms cancel, leaving 2a — the length of the major axis. So the constant in the definition is always the full major axis length.
Sorting
The major axis is longer and holds the foci.
Sort into buckets
Sort each feature.
Since a is always greater than b, the axis with the larger value is always the major one — which is why the equation's larger denominator identifies it.
Prediction
Commit before reasoning.
Predict first
Move the two foci of an ellipse together until they coincide. What shape results?
Correct: A circle, since c becomes 0 and a equals b.
\[ c = 0 \;\Longrightarrow\; a = b \;\Longrightarrow\; x^2+y^2 = a^2 \]
Why: With c equal to 0 the relation c squared equals a squared minus b squared forces a to equal b, so both denominators are the same and the equation becomes x squared plus y squared equals a squared — a circle of radius a. Physically the string is now pinned at one point and the pencil stays a fixed distance from it. So a circle is an ellipse whose two foci have merged, which is why the two definitions look so similar.
Section
Section 2
Concept
An ellipse centred at the origin has x squared over a squared plus y squared over b squared equal to 1 when the major axis is horizontal, and the denominators exchanged when it is vertical. Always a is greater than b.
\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad \text{or} \quad \frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \]
Because a is always the larger, the orientation is decided by which denominator is bigger — under x means horizontal, under y means vertical.
Figure (svg): The parts of an ellipse labelled, in both the horizontal and vertical orientations
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-634 — Standard Equation of an Ellipse with Center at the Origin
Picture it
The two standard ellipses with their parts labelled.
Figure (svg): The parts of an ellipse labelled, in both the horizontal and vertical orientations
On the left the larger denominator sits under x and the foci lie left and right; on the right it sits under y and the foci lie above and below.
Worked example
Applying the rule to Example 1 and the guided practice.
\[ \text{Which axis is major for } \tfrac{x^2}{25}+\tfrac{y^2}{4}=1, \; \tfrac{x^2}{16}+\tfrac{y^2}{9}=1, \; \tfrac{x^2}{36}+\tfrac{y^2}{49}=1, \; \tfrac{x^2}{9}+\tfrac{y^2}{25}=1? \]
First: compare 25 and 4
Why: The larger sits under x.
Second: compare 16 and 9
Why: The larger sits under x again.
Third: compare 36 and 49
Why: The larger sits under y.
Fourth: compare 9 and 25
Why: The larger sits under y.
Figure (svg): The parts of an ellipse labelled, in both the horizontal and vertical orientations
\[ H, \; H, \; V, \; V \]
Verify: check a is always the larger
Why: In every case a came out greater than b, as the definition requires. If a computation ever gave a smaller than b, the two would simply have been mislabelled — the letter a always names the larger semi-axis, whichever variable it sits under.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-635
Fill the middle
Example 1.
Fill in the blanks
4x^2+25y^2 = 100 \;\Longrightarrow\; \frac25___}+\frac______ = 1
Why: One hundred divided by 4 is 25, so a squared is 25 and a is 5. Note how a coefficient of 4 in front becomes a denominator of 25 underneath.
Worked example
Example 1 Step 1 and Guided Practice 3.
\[ \text{Put } 4x^2+25y^2=100 \text{ and } 25x^2+9y^2=225 \text{ into standard form.} \]
First: divide by the constant
Why: Every term over 100.
\[ x ^{2} / 25 + y ^{2} / 4 = 1 \]
First: read the denominators
Why: Twenty-five and 4.
\[ a = 5, b = 2 \]
Second: divide by 225
Why: Two hundred twenty-five over 25 is 9, and over 9 is 25.
\[ x ^{2} / 9 + y ^{2} / 25 = 1 \]
Second: read the denominators
Why: The larger is under y.
Figure (svg): The solution to Worked example rewrite in standard form shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{x^2}{25}+\tfrac{y^2}{4}=1; \quad \tfrac{x^2}{9}+\tfrac{y^2}{25}=1 \]
Verify: check a vertex in each
Why: For the first, 5 comma 0 gives 25 over 25 plus 0, which is 1. For the second, 0 comma 5 gives 0 plus 25 over 25, also 1. Testing a vertex confirms both the division and the reading of which denominator is which.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-635
Error analysis
A student identifies the semi-axes from an equation not yet in standard form.
Annotate
On: \( 4x^2+25y^2 = 100 \;\Longrightarrow\; a^2 = 4, \; b^2 = 25 \)
A coefficient in front and a denominator underneath are reciprocals of each other, so reading one as the other reverses the whole orientation.
Sorting
Find the larger denominator.
Sort into buckets
Sort each ellipse by its major axis.
One comparison decides it. The numbers themselves do not matter — only which of the two is bigger.
Comparison
Fill the blanks. Swap the denominators.
Comparison matrix
| Feature | Horizontal major axis | Vertical major axis |
|---|---|---|
| Equation | x^2/a^2 + y^2/b^2 = 1 | x^2/b^2 + y^2/a^2 = 1 |
| Vertices | (+-a, 0) | (0, +-a) |
| Co-vertices | (0, +-b) | (+-b, 0) |
| Foci | (+-c, 0) | (0, +-c) |
Every entry swaps its coordinates between the columns. There is really one form, seen from two directions.
Prediction
Commit before reasoning.
Predict first
The circle's equation had r squared on the right. Why does the ellipse's have 1?
Correct: So that both semi-axes can be read directly as denominators, since one number cannot carry two lengths.
\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text{ carries both } a \text{ and } b \]
Why: A circle has a single radius, so one constant on the right suffices. An ellipse has two different semi-axes, so each needs its own place — and putting them under the squares with 1 on the right does exactly that. The same convention will carry into the hyperbola of Lesson 9.5, where again two lengths must be recorded separately.
Section
Section 3
Concept
Divide into standard form, take square roots of both denominators to get a and b, plot the four axis points, and sketch. The foci follow from c squared equal to a squared minus b squared.
\[ c^2 = a^2-b^2 \]
The subtraction always goes this way round, larger minus smaller, so c comes out real and less than a. That places both foci inside the curve.
Figure (svg): An ellipse graphed from a rearranged equation, with vertices, co-vertices and foci marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-635 — Graph an equation of an ellipse
Picture it
Example 1, fully labelled.
Figure (svg): An ellipse graphed from a rearranged equation, with vertices, co-vertices and foci marked
The vertices are 5 units out horizontally, the co-vertices 2 units vertically, and the foci about 4.6 units along the major axis.
Worked example
Example 1.
\[ \text{Graph } 4x^2+25y^2=100 \text{ and identify the vertices, co-vertices and foci.} \]
Divide into standard form
Why: Every term over 100.
\[ x ^{2} / 25 + y ^{2} / 4 = 1 \]
Find a and b
Why: The square roots of 25 and 4.
\[ a = 5, b = 2 \]
Read the orientation and the points
Why: Larger denominator under x.
\[ (+- 5, 0)\text{ and } (0, +- 2) \]
Find c
Why: Twenty-five minus 4 is 21.
\[ c = \sqrt{21}\text{ about } 4.6 \]
Figure (svg): An ellipse graphed from a rearranged equation, with vertices, co-vertices and foci marked
\[ (\pm 5,0), \; (0,\pm 2), \; (\pm\sqrt{21},0) \]
Verify: check the constant sum at a co-vertex
Why: From 0 comma 2 to each focus is the square root of 21 plus 4, which is 5. The two distances add to 10, which is 2a — exactly as the definition promises. Checking the definition at a co-vertex tests a, b and c all at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-635
Fill the middle
Example 1.
Fill in the blanks
c^2 = 25-4 = 21
Why: Twenty-five minus 4 is 21, so c is the square root of 21, about 4.6. That is less than a, which places both foci inside the ellipse.
Worked example
Guided Practice 1 to 3.
\[ \text{Identify the parts of } \tfrac{x^2}{16}+\tfrac{y^2}{9}=1, \; \tfrac{x^2}{36}+\tfrac{y^2}{49}=1, \; 25x^2+9y^2=225. \]
First: horizontal, a is 4 and b is 3
Why: Sixteen minus 9 is 7.
\[ (+- 4, 0), (0, +- 3), (+- \sqrt{7}, 0) \]
Second: vertical, a is 7 and b is 6
Why: Forty-nine minus 36 is 13.
\[ (0, +- 7), (+- 6, 0), (0, +- \sqrt{13}) \]
Third: divide by 225
Why: X squared over 9 plus y squared over 25.
Third: find c
Why: Twenty-five minus 9 is 16.
\[ c = 4,\text{ foci } (0, +- 4) \]
Figure (svg): The solution to Worked example three more ellipses shown as a ladder of expressions, one row per algebraic move
\[ (\pm\sqrt7,0); \; (0,\pm\sqrt{13}); \; (0,\pm 4) \]
Verify: check that each c is less than its a
Why: The three values of c are about 2.65, 3.61 and 4, against a values of 4, 7 and 5. Every focus lies comfortably inside its ellipse, as it must. A c exceeding a would mean the subtraction had been done the wrong way round.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-635
Trap
\[ \frac{x^2}{25}+\frac{y^2}{4}=1 \]
Compute c squared as b squared minus a squared
Why: The subtraction is written from the smaller number first.
\[ c^2 = 4-25 = -21 \quad \text{(no real } c\text{)} \]
A negative square would leave no real foci at all, which cannot be right — every ellipse has two.
\[ c^2 = a^2-b^2 = 25-4 = 21 \]
Always subtract the smaller from the larger
Why: Since a exceeds b, this order always gives a positive result.
\[ c = \sqrt{21} \approx 4.6 < 5 \quad \checkmark \]
A negative value under the root is an immediate signal that a and b have been swapped. In Lesson 9.5 the hyperbola will use a PLUS in this relation instead, so the direction matters.
Matching
Subtract, then root.
Match the pairs
Why: The last two have their foci on the vertical axis because their larger denominators sit under y. Only the fourth gives a whole-number c, since 25 minus 9 is 16.
Ranking
Graphing an ellipse from any equation.
Put in order
Why: Step one is what makes the denominators mean anything, and step three is what tells you where the foci go in step five. Skipping the division reverses the orientation, since coefficients and denominators are reciprocal.
Prediction
Commit before reasoning.
Predict first
Example 1 has a equal to 5 and c about 4.6. What does the closeness of c to a mean?
Correct: The ellipse is very elongated, since the foci sit near the vertices.
\[ e = \frac{c}{a} = \frac{4.58}{5} \approx 0.92 \]
Why: The ratio c over a is called the eccentricity, and here it is about 0.92 — close to 1, which means a long thin ellipse. The semi-axes confirm it: 5 across against only 2 up. When c is near zero the foci huddle at the centre and the ellipse is nearly a circle, which is the case for most planetary orbits. Earth's eccentricity is about 0.017, which is why its orbit looks circular in every diagram.
Section
Section 4
Concept
A vertex gives a and tells you which axis is major. A co-vertex then gives b directly; a focus gives c instead, and b follows from c squared equal to a squared minus b squared.
\[ a=8, \; c=4 \;\Longrightarrow\; b^2 = 64-16 = 48 \]
Only b squared is ever needed, so a radical value of b never has to be simplified before it goes into the equation.
Figure (svg): Two ellipses built from partial information, one from a co-vertex and one from a focus
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-636 — Write an equation given a vertex and a co-vertex
Picture it
Examples 2 and 4, built from different second facts.
Figure (svg): Two ellipses built from partial information, one from a co-vertex and one from a focus
On the left the co-vertex gave b equal to 3 at once. On the right the focus gave c equal to 4, and b squared came out as 48 by subtraction.
Worked example
Example 2.
\[ \text{Write the ellipse with vertex } (0,4), \text{ co-vertex } (-3,0) \text{ and centre } (0,0). \]
Read the orientation
Why: The vertex is on the vertical axis.
Read a and b
Why: Four from the vertex, 3 from the co-vertex.
\[ a = 4, b = 3 \]
Choose the form
Why: Larger denominator under y.
\[ x ^{2} / b ^{2} + y ^{2} / a ^{2} = 1 \]
Substitute
Why: Nine under x and 16 under y.
\[ x ^{2} / 9 + y ^{2} / 16 = 1 \]
Figure (svg): Two ellipses built from partial information, one from a co-vertex and one from a focus
\[ \frac{x^2}{9}+\frac{y^2}{16}=1 \]
Verify: test both given points
Why: At 0 comma 4 the equation gives 0 plus 16 over 16, which is 1. At negative 3 comma 0 it gives 9 over 9 plus 0, also 1. Both lie on the curve, and by symmetry so do 0 comma negative 4 and 3 comma 0.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 635-635
Fill the middle
Example 4.
Fill in the blanks
c^2 = a^2-b^2: \; 16 = 64-b^2 \;\Longrightarrow\; b^2 = 48
Why: Sixty-four minus 16 is 48, which goes straight into the equation as a denominator. There is no need to simplify root 48, since only b squared appears.
Worked example
Example 4 and Guided Practice 6 and 7.
\[ \text{Write the ellipses with vertex } (-8,0) \text{ and focus } (4,0); \text{ vertex } (0,8) \text{ and focus } (0,-3). \]
First: read a and c
Why: Both lie on the horizontal axis.
\[ a = 8, c = 4 \]
First: find b squared
Why: Sixty-four minus 16.
\[ b ^{2} = 48 \]
First: write the equation
Why: Larger denominator under x.
\[ x ^{2} / 64 + y ^{2} / 48 = 1 \]
Second: a is 8 and c is 3, vertical
Why: Sixty-four minus 9 is 55.
\[ x ^{2} / 55 + y ^{2} / 64 = 1 \]
Figure (svg): Two ellipses built from partial information, one from a co-vertex and one from a focus
\[ \tfrac{x^2}{64}+\tfrac{y^2}{48}=1; \quad \tfrac{x^2}{55}+\tfrac{y^2}{64}=1 \]
Verify: check b is smaller than a in both
Why: Forty-eight is less than 64, and 55 is less than 64, so in both the co-vertex denominator is the smaller one — consistent with the stated orientations. Note that b itself is irrational in both cases, root 48 and root 55, but only b squared appears in the equation, so neither radical ever has to be written.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 636-636
Error analysis
A student writes an ellipse with vertex at 8 units and a focus at 4 units, both on the horizontal axis.
Annotate
On: \( a = 8, \; b = 4 \;\Longrightarrow\; \frac{x^2}{64}+\frac{y^2}{16}=1 \)
A focus and a co-vertex are never on the same axis, so their positions distinguish them. Reading which axis the given point sits on settles which letter it supplies.
Sorting
A vertex gives a; a co-vertex gives b; a focus gives c.
Sort into buckets
Sort each given point for an ellipse centred at the origin.
The vertex and the focus share an axis, and the co-vertex is on the other one — which is how a picture tells them apart even without labels.
Matching
Vertex first, then the second fact.
Match the pairs
Why: The first two needed no subtraction at all, since a co-vertex gives b directly. The last two each cost one subtraction to turn c into b squared.
Prediction
Commit before reasoning.
Predict first
Why is a vertex the most useful single clue about an ellipse?
Correct: Because it gives a and also reveals which axis is major.
\[ \text{vertex } (0,a) \;\Longrightarrow\; \text{vertical major axis and the value of } a \]
Why: A vertex is on the major axis by definition, so where it sits settles the orientation, and its distance from the centre is a. That is two facts from one point. A co-vertex would also reveal the orientation, but a focus alone would not tell you a — you would know c and still need more. This is why every problem in the lesson supplies a vertex plus exactly one other thing.
Section
Section 5
Concept
A region described by its full width and height gives a and b after halving each. The area enclosed by an ellipse is pi times a times b, which reduces to pi r squared when the two are equal.
\[ A = \pi ab \]
The stated dimensions are always full widths, so halving comes first. Forgetting it makes the area four times too large.
Figure (svg): An elliptical region where a lightning strike most likely landed, with its dimensions and area
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 636-636 — Multi-step problem
Picture it
Example 3: an elliptical region 400 metres by 200 metres.
Figure (svg): An elliptical region where a lightning strike most likely landed, with its dimensions and area
Halving gives 200 and 100, so the equation has denominators 40,000 and 10,000. The area is about 62,800 square metres.
Worked example
Example 3.
\[ \text{A region is } 400 \text{ m east to west and } 200 \text{ m north to south. Find its equation and area.} \]
Halve both dimensions
Why: Four hundred over 2 and 200 over 2.
\[ a = 200, b = 100 \]
Identify the orientation
Why: The horizontal extent is larger.
Write the equation
Why: Two hundred squared and 100 squared.
\[ x ^{2} / 40, 000 + y ^{2} / 10, 000 = 1 \]
Find the area
Why: Pi times 200 times 100.
\[ \text{about } 62, 800\text{ square metres} \]
Figure (svg): An elliptical region where a lightning strike most likely landed, with its dimensions and area
\[ A = \pi(200)(100) \approx 62{,}800 \text{ m}^2 \]
Verify: compare with a rectangle
Why: The bounding rectangle is 400 by 200, or 80,000 square metres, and the ellipse covers about 62,800 — roughly 78.5 percent of it. That ratio is pi over 4 and is the same for every ellipse, which is a useful sanity check on any area computed this way.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 636-636
Fill the middle
Example 3.
Fill in the blanks
a = \frac200___ = ___
Why: Four hundred metres is the full width, so the semi-axis is 200. Using 400 would make the area four times too large.
Worked example
Guided Practice 8.
\[ \text{Repeat for a region } 250 \text{ m east to west and } 350 \text{ m north to south.} \]
Halve both dimensions
Why: One hundred twenty-five and 175.
\[ b = 125, a = 175 \]
Identify the orientation
Why: The vertical extent is now larger.
Write the equation
Why: One hundred twenty-five squared and 175 squared.
\[ x ^{2} / 15, 625 + y ^{2} / 30, 625 = 1 \]
Find the area
Why: Pi times 175 times 125.
\[ \text{about } 68, 700\text{ square metres} \]
Figure (svg): The solution to Worked example a taller region shown as a ladder of expressions, one row per algebraic move
\[ A = \pi(175)(125) \approx 68{,}700 \text{ m}^2 \]
Verify: check the orientation against the equation
Why: The larger denominator, 30,625, sits under y squared, matching a region taller than it is wide. And the area is again about 78.5 percent of the 250 by 350 bounding rectangle, which is 87,500. Both checks confirm the halving was done and the letters assigned correctly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 636-636
Trap
\[ \text{region } 400 \text{ by } 200 \text{ metres} \]
Take a as 400 and b as 200
Why: The stated dimensions are used directly.
\[ A = \pi(400)(200) \approx 251{,}000 \quad \text{(wrong)} \]
That is larger than the whole bounding rectangle of 80,000 square metres, which is impossible for a region inside it.
\[ a = 200, \; b = 100 \;\Longrightarrow\; A = \pi(200)(100) \approx 62{,}800 \]
Halve each stated dimension first
Why: The letters a and b are SEMI-axes, measured from the centre.
\[ 62{,}800 < 80{,}000 \quad \checkmark \]
The same halving was needed for the parabolic dish in Lesson 9.2, and comparing against the bounding rectangle catches the error either time.
Comparison
Fill the blanks. Different shapes, similar areas.
Comparison matrix
| Quantity | 400 by 200 m | 250 by 350 m |
|---|---|---|
| a and b | 200 and 100 | 175 and 125 |
| Major axis | horizontal | vertical |
| Area | about 62,800 square metres | about 68,700 square metres |
| Shape | long and thin | closer to circular |
The second region has a smaller bounding rectangle in one direction yet a larger area, because area depends on the product of the semi-axes rather than on either alone.
Prediction
Commit before reasoning.
Predict first
What does the area formula pi a b become when a equals b?
Correct: Pi r squared, the circle's area, since a and b both equal the radius.
\[ a = b = r \;\Longrightarrow\; A = \pi r^2 \]
Why: When the two foci merge, a equals b equals r, and pi a b becomes pi r squared. So the ellipse formula contains the circle formula as its special case, exactly as the ellipse's definition contains the circle's. Reading a general formula's special cases is a good habit: it checks the formula and it links the new fact to one you already trust.
Ranking
From a region's dimensions to its area.
Put in order
Why: Step two is where the errors happen and step three is where the orientation is fixed. Note that step five does not depend on which is which, since the product is the same either way — but the equation in step four certainly does.
Comparison
Fill the blanks. Each is a statement about distance.
Comparison matrix
| Curve | Distance condition | Equation at the origin |
|---|---|---|
| Parabola | equal to a point and a line | x^2 = 4py |
| Circle | fixed distance from one point | x^2 + y^2 = r^2 |
| Ellipse | fixed SUM of distances to two points | x^2/a^2 + y^2/b^2 = 1 |
| Circle as an ellipse | the two foci coincide | a = b, so c = 0 |
Each new curve loosens the previous condition by one degree, and the equations grow to hold the extra information that loosening requires.
Pattern
One routine to graph, one to write, one to model.
The foci always lie on the major axis and always inside the curve, so c is always less than a. A c larger than a means the letters were swapped.
OpenStax Algebra and Trigonometry 2e, §12.1 The Ellipse §12.1
Check
Standard form first.
Check your understanding
What are the vertices of 4x^2 + 25y^2 = 100?
Answer: A
Why: Dividing by 100 gives x^2/25 + y^2/4 = 1, so a = 5 on the horizontal axis.
Check
Foci. Larger minus smaller.
Check your understanding
Where are the foci of x^2/9 + y^2/25 = 1?
Answer: A
Why: The major axis is vertical, and c^2 = 25 - 9 = 16.
Check
A vertex and a focus.
Check your understanding
Write the ellipse with vertex (-8, 0), focus (4, 0) and centre (0, 0).
Answer: A
Why: The vertex gives a = 8 and the focus gives c = 4, so b^2 = 64 - 16 = 48.
Real world
Every planet's orbit is an ellipse with the sun at one focus. Earth's orbit has a semi-major axis of about 149.6 million kilometres and an eccentricity, c over a, of about 0.0167.
Discussion prompt
Find c and b for Earth's orbit, and say how far the orbit departs from a circle.
Hint: Get c from the eccentricity, then b from c squared equals a squared minus b squared.
Answer:
\[ c = 0.0167a \approx 2.50 \text{ million km} \]
\[ b^2 = a^2-c^2 \;\Longrightarrow\; b \approx \sqrt{149.6^2-2.50^2} \approx 149.58 \]
The semi-minor axis is about 149.58 million kilometres, against a semi-major axis of 149.6 — a difference of about two hundredths of one percent. Drawn to scale, Earth's orbit is indistinguishable from a circle.
What is not negligible is the sun's position: it sits at a focus, 2.5 million kilometres off centre, so Earth is about 5 million kilometres closer to it in January than in July. That is a 3 percent swing in distance, which changes the sunlight received by about 7 percent — and yet it is not what causes the seasons, since January is winter in the northern hemisphere. The axial tilt does that. Kepler's first law, that orbits are ellipses with the sun at a focus, was exactly this lesson's definition applied to astronomical data, and it took him years precisely because the departure from a circle is so slight.
Commit first
Answer, then rate your confidence honestly.
Predict first
Do the foci of an ellipse lie on the minor axis?
Correct: No — they lie on the major axis, and always inside the curve.
\[ c^2 = a^2-b^2 < a^2 \;\Longrightarrow\; c < a \]
Why: The definition builds the ellipse from two foci, and the line through them meets the curve at the vertices — so that line is the major axis by construction. The algebra agrees: c squared equals a squared minus b squared makes c less than a, so the foci sit between the centre and the vertices. Placing them on the minor axis would put them at a distance of about 4.6 in Example 1, where the ellipse only reaches 2 — outside the curve entirely, which is impossible for points the curve is defined by.
Explain it
They know what a circle is and think an ellipse is just a squashed one.
Discussion prompt
In four sentences or fewer, explain what makes a curve an ellipse rather than any old oval.
Hint: Describe the string-and-tacks construction.
Answer:
Push two tacks into a board and loop a piece of string loosely around them. Put a pencil inside the loop, pull it taut, and draw all the way round.
The curve you get is an ellipse, because at every moment the two distances from the pencil to the tacks add up to the same total — the string's length. Move the tacks together and you get a circle; move them apart and the ellipse gets longer and thinner.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For dividing, put a 1 on the right before reading anything. For the major axis, circle the larger denominator. For c, always write larger minus smaller and check that c comes out less than a. For focus versus co-vertex, ask which axis the point sits on — a focus shares its axis with the vertices.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an ellipse page. Top left: draw an ellipse with both foci marked, pick two points on it, and verify that both pairs of distances add to the same total, writing that total as 2a. Top right: draw both standard orientations side by side with every part labelled, and write the rule for reading the orientation from the equation. Middle: work Example 1 in full, from dividing through to marking the foci, and check the constant sum at a co-vertex. Bottom left: work Examples 2 and 4, noting beside each which letter the second clue supplied. Bottom right: draw the lightning region, mark the halving of both dimensions, compute the area, and compare it with the bounding rectangle as a check.
If any of your foci lie outside the curve, recheck the subtraction: c squared is a squared minus b squared, larger minus smaller, and c must come out less than a.
Recap
Five things, and a curve that contains the circle as a special case.
| If you see | Then |
|---|---|
| Two squares with different denominators, summing to 1 | An ellipse |
| The larger denominator under x squared | A horizontal major axis |
| A vertex | It gives a and the orientation |
| A co-vertex | It gives b directly |
| A focus | It gives c; then b squared is a squared minus c squared |
| A region's full width | Halve it before using it as a semi-axis |
Lesson 9.5 changes the plus between the two squares into a minus, which turns the closed ellipse into the two open branches of a hyperbola.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.4 Graph and Write Equations of Ellipses §9.4, pp. 634-637 — everything on these slides traces back here
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