Changes the ellipse's constant sum to a constant difference and follows the consequences: two unbounded branches, foci outside the curve, a rearranged relation among the constants, and asymptotes found from a central rectangle. Covers both orientations and translated hyperbolas.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 10 — Analytic Geometry
§10.2 The Hyperbola, pp. 1208-1229
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1208-1229 — the pages these objectives are drawn from
Warm-up
One word of the ellipse's definition is about to change.
Discussion prompt
Instead of a constant sum of distances to two fixed points, suppose the difference is constant. What changes?
Hint: Can the point wander far away?
Answer:
A constant sum bounds how far the point can go — neither distance can exceed the total. So the ellipse closes.
A constant difference imposes no such bound. Both distances can grow indefinitely as long as their difference stays fixed.
So the curve runs off to infinity, in two branches — one where the first distance is larger and one where the second is. That is the hyperbola.
Concept
A hyperbola is the set of points whose distances to two fixed foci differ by a constant amount. Because nothing is bounded, the curve consists of two unbounded branches.
hyperbola — the set of all points whose distances to two fixed points differ by a constant amount
\[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad c^2=a^2+b^2 \]
The minus sign in the equation is the visible trace of the minus sign in the definition. Everything else about the section follows from the curve being unbounded, which is what the difference permits.
Figure (svg): A hyperbola with two branches, its foci marked outside the branches, and a point joined to both foci
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1208-1213
Section
Section 1
Concept
Requiring a constant difference rather than a constant sum removes the bound on how far a point can stray, so the curve opens into two unbounded branches.
Checking the constant at a vertex works exactly as it did for the ellipse. From the near vertex the two distances are the focal distance minus and plus the semi-transverse axis, and their difference is twice the semi-transverse axis.
Figure (svg): A hyperbola with two branches, its foci marked outside the branches, and a point joined to both foci
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1208-1214
Picture it
Two distances with a constant difference.
Figure (svg): A hyperbola with two branches, its foci marked outside the branches, and a point joined to both foci
The dashed lines are asymptotes, which an ellipse has no analogue for. They exist because the branches are unbounded and have to approach some direction.
Worked example
Check at a vertex.
\[ \text{A hyperbola has vertices at } (\pm 3,0) \text{ and foci at } (\pm 5,0). \text{ Find the constant difference.} \]
Go to the right vertex
Why: At three units out.
Find the near distance
Why: To the right focus.
\[ 5 - 3 = 2 \]
Find the far distance
Why: To the left focus.
\[ 5 + 3 = 8 \]
Subtract
Why: The constant difference.
\[ 6 \]
Figure (svg): A hyperbola with two branches, its foci marked outside the branches, and a point joined to both foci
\[ \text{difference}=2a=6 \]
Verify: check the other vertex
Why: At the left vertex the two distances swap, giving 8 and 2 — the same difference of 6 with the opposite sign. Taking the difference as a magnitude covers both branches, which is why the definition uses the absolute difference.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1209-1211
Sorting
One word of the definition separates them.
Sort into buckets
Sort each property.
Worked example
The sign of the difference distinguishes them.
\[ \text{Explain why a constant difference gives two branches.} \]
Consider the first case
Why: The left distance exceeds the right.
Consider the second
Why: The right exceeds the left.
Note both satisfy the definition
Why: Same magnitude of difference.
Conclude
Why: Two separate curves.
Figure (svg): The solution to Worked example why two branches shown as a ladder of expressions, one row per legal move
\[ \text{two branches, one per sign} \]
Verify: compare with the ellipse
Why: A sum has no sign choice — two distances always add to a positive total — so the ellipse is a single curve. The difference's sign is what doubles the hyperbola, which is a genuine structural consequence of the one changed word.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1211-1214
Trap
\[ \text{a hyperbola is a stretched ellipse} \]
Treat it as a variant of the same closed shape
Why: The change from sum to difference is read as a change of proportion.
The unbounded branches and the asymptotes have no place in that picture.
A constant sum bounds the curve and a constant difference does not. They are different kinds of object, not different proportions.
The hyperbola's branches run off indefinitely, approaching asymptotes that an ellipse has no analogue for.
The foci also move outside the curve, one inside each branch, where an ellipse's foci are inside the single closed loop.
Prediction
A hyperbola's foci relative to its branches.
Predict first
Where are they?
Correct: Outside the branches, one inside each opening.
Why: The focal distance exceeds the semi-transverse axis for a hyperbola, so each focus lies beyond its vertex, inside the opening of that branch. For an ellipse the reverse holds and the foci are inside the closed curve.
Faded example
With semi-transverse axis 3 and focal distance 5.
Fill in the blanks
(5+3)-(5-3)=8-2=6
Why: At a vertex the two distances are the focal distance plus and minus the semi-transverse axis, so subtracting cancels the focal distance and leaves twice the semi-transverse axis.
Explain it to yourself
A hyperbola comes in two pieces.
Discussion prompt
Explain why, using the definition.
Hint: What sign can a difference have?
Answer:
A difference can be positive or negative with the same magnitude, depending on which distance is larger. Both cases satisfy the definition.
Points where the first focus is further form one branch; points where the second is further form the other. They are separate because no point can satisfy both.
A sum has no such choice, since two distances always add to one positive total. That is why the ellipse is a single curve and the hyperbola is two, which a good explanation traces back to the one changed word.
Section
Section 2
Concept
A hyperbola opens along the axis of whichever variable carries the positive term. The relative size of the denominators is irrelevant here, unlike for an ellipse.
Carrying over the ellipse's rule is the standard error here: there the larger denominator decided the orientation, and here the sign does. A hyperbola can perfectly well have a larger denominator under the negative term.
Figure (svg): A card showing that the positive term identifies the axis along which the hyperbola opens
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1214-1218
Picture it
The sign decides, not the size.
Figure (svg): A card showing that the positive term identifies the axis along which the hyperbola opens
The red line names the trap directly. For an ellipse the larger denominator was the deciding feature, and importing that habit here gives the wrong axis whenever the denominators happen to be ordered the other way.
Worked example
Find the positive term.
\[ \text{Describe } \frac{y^2}{9}-\frac{x^2}{16}=1. \]
Find the positive term
Why: The vertical one.
Read the semi-transverse axis
Why: Root of its denominator.
\[ a = 3 \]
Read the other constant
Why: Root of the other.
\[ b = 4 \]
Locate the vertices
Why: On the vertical axis.
Figure (svg): A card showing that the positive term identifies the axis along which the hyperbola opens
\[ \text{opens vertically},\; a=3,\; b=4 \]
Verify: check the denominator sizes
Why: Here the larger denominator, 16, sits under the negative term — and the hyperbola still opens vertically. Applying the ellipse's rule would have given the wrong axis, which is exactly the trap this example exposes.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1215-1216
Prediction
The vertical term is positive and has the smaller denominator.
Predict first
Which way does the hyperbola open?
Correct: Vertically, since that term is positive.
Why: The sign decides the orientation for a hyperbola, not the size of the denominators. Importing the ellipse's rule gives the wrong axis whenever the denominators happen to be ordered the other way.
Worked example
The signs have swapped.
\[ \text{Describe } \frac{x^2}{16}-\frac{y^2}{9}=1. \]
Find the positive term
Why: The horizontal one.
Read the semi-transverse axis
Why: Root of its denominator.
\[ a = 4 \]
Read the other constant
Why: Root of the other.
\[ b = 3 \]
Locate the vertices
Why: On the horizontal axis.
Figure (svg): The solution to Worked example the other orientation shown as a ladder of expressions, one row per legal move
\[ \text{opens horizontally},\; a=4,\; b=3 \]
Verify: compare with the previous example
Why: The two equations use the same numbers and differ only in which term is positive — and the two hyperbolas open along perpendicular axes. The sign, and nothing else, made the difference.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1216-1218
Error analysis
A student reads a hyperbola's orientation.
Annotate
On: \( \frac{y^2}{9}-\frac{x^2}{16}=1 \;\Longrightarrow\; \text{opens horizontally, since }16>9 \)
The two conics use different rules for the same question, which is why importing one habit into the other section gives a plausible but wrong answer. Checking which term is positive takes one glance.
Sorting
Find the positive term.
Sort into buckets
Sort each equation.
Faded example
From a hyperbola whose vertical term is positive with denominator 9.
Fill in the blanks
a=\sqrt9}=3, \text___
Why: The semi-transverse axis comes from the positive term's denominator, whatever its size relative to the other. Its square root gives the distance from the centre to each vertex.
Step zero
You are given a hyperbola's equation to describe.
Discussion prompt
What do you look at first?
Hint: Not the sizes.
Answer:
Which term is positive. That names the axis the branches open along and identifies which denominator gives the semi-transverse axis.
Comparing the denominators — the ellipse's first move — settles nothing here and can actively mislead.
So the two conics ask different first questions. A good habit is to name the conic before reading anything else, since that determines which rule applies.
Section
Section 3
Concept
For a hyperbola the relation among the three constants puts the focal distance alone on one side, so it is the largest — which is why the foci lie beyond the vertices.
The consequence is geometric: because the focal distance exceeds the semi-transverse axis, each focus lies beyond its vertex — outside the curve. For an ellipse the reverse holds and the foci are enclosed.
| ellipse | hyperbola | |
|---|---|---|
| relation | a squared equals b squared plus c squared | c squared equals a squared plus b squared |
| hypotenuse | the semi-major axis | the focal distance |
| largest constant | a | c |
| foci relative to vertices | closer to the centre | further from the centre |
Figure (svg): A contrast between the ellipse's relation among its three constants and the hyperbola's
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1218-1222
Picture it
The same three letters, rearranged.
Figure (svg): A contrast between the ellipse's relation among its three constants and the hyperbola's
The last row on each side is the visible consequence. Which constant is the hypotenuse decides whether the foci end up inside a closed curve or beyond the vertices of an open one.
Worked example
Add, rather than subtract.
\[ \text{Find the foci of } \frac{x^2}{16}-\frac{y^2}{9}=1. \]
Identify the two squares
Why: From the denominators.
\[ 16\text{ and } 9 \]
Apply the relation
Why: Add them.
\[ c ^{2} = 25 \]
Take the root
Why: The focal distance.
\[ c = 5 \]
Place them
Why: On the axis of opening.
Figure (svg): A contrast between the ellipse's relation among its three constants and the hyperbola's
\[ (\pm 5,0) \]
Verify: check the focus lies beyond the vertex
Why: The vertex is at 4 and the focus at 5, so the focus is outside the curve — which is required, since a hyperbola's foci sit inside the openings of its branches. For an ellipse the focus would have been inside instead.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1219-1221
Prediction
For a hyperbola.
Predict first
Which of the three is the largest?
Correct: The focal distance, being the hypotenuse.
Why: The hyperbola's relation puts the focal distance alone on one side as the hypotenuse, so it exceeds both others. That is why its foci lie beyond the vertices rather than inside a closed curve.
Worked example
The relation runs both ways.
\[ \text{Find the equation with vertices at } (\pm 3,0) \text{ and foci at } (\pm 5,0). \]
Read the semi-transverse axis
Why: From the vertices.
\[ a = 3 \]
Read the focal distance
Why: From the foci.
\[ c = 5 \]
Apply the relation
Why: Subtract to find the other.
\[ b ^{2} = 25 - 9 = 16 \]
Write the equation
Why: Positive term on the axis of opening.
\[ x ^{2} / 9 - y ^{2} / 16 = 1 \]
Figure (svg): The solution to Worked example build from the foci shown as a ladder of expressions, one row per legal move
\[ \frac{x^2}{9}-\frac{y^2}{16}=1 \]
Verify: check the denominator sizes
Why: Here the larger denominator ends up under the negative term, which is perfectly legitimate for a hyperbola. The construction was driven by the relation and the orientation, and the size ordering fell out however it did.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1221-1222
Trap
\[ c^2=a^2-b^2=16-9=7 \]
Subtract, as for an ellipse
Why: The relation is recalled from the previous section.
The focal distance comes out less than the semi-transverse axis, putting the foci inside the branches where no point of the curve is.
For a hyperbola the focal distance is the hypotenuse, so its square is the sum of the other two.
That makes the focal distance the largest, which is why the foci lie beyond the vertices.
Check that the focal distance exceeds the semi-transverse axis. A focus inside a branch is impossible and the check catches the wrong relation immediately.
Faded example
With denominators 16 and 9.
Fill in the blanks
c^2=16+9=25 \;\Longrightarrow\; c=5
Why: The hyperbola's relation adds the two squares, unlike the ellipse's which subtracts. The resulting focal distance exceeds both constants, placing the foci outside the branches.
Sorting
The conic decides.
Sort into buckets
Sort each statement.
Explain it
An ellipse encloses its foci and a hyperbola does not.
Discussion prompt
Explain to a classmate what causes the difference.
Hint: Which constant is largest in each?
Answer:
For an ellipse, the semi-major axis is the hypotenuse, so it exceeds the focal distance — each focus is closer to the centre than the vertex is, and therefore inside.
For a hyperbola, the focal distance is the hypotenuse, so it exceeds the semi-transverse axis — each focus is further out than its vertex, and therefore outside the branch.
So the rearranged relation is not a bookkeeping detail. It determines where the foci sit relative to the curve, which a good explanation points out is the visible difference between the two conics.
Section
Section 4
Concept
The branches approach two straight lines through the centre. Drawing a rectangle extending a along the axis of opening and b perpendicular to it gives those lines as its diagonals.
The rectangle is a construction rather than part of the curve, but it is the fastest route to an accurate sketch. Drawing it first fixes the vertices, the asymptote slopes and the scale in one step.
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running along the rectangle's diagonals
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1222-1226
Picture it
Its diagonals are the asymptotes.
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running along the rectangle's diagonals
The branches touch the rectangle at the vertices and then bend away towards the diagonals. Sketching in that order — rectangle, diagonals, branches — gives a correct picture without computing any points.
Worked example
Slopes from the rectangle's corners.
\[ \text{Find the asymptotes of } \frac{x^2}{16}-\frac{y^2}{9}=1. \]
Read the two constants
Why: Square roots of the denominators.
\[ a = 4, b = 3 \]
Build the rectangle
Why: Four either side, three up and down.
\[ 8\text{ by } 6 \]
Find the diagonal slopes
Why: Rise over run to a corner.
\[ \frac{3}{4}\text{ and } -\frac{3}{4} \]
Write the lines
Why: Through the centre.
\[ y =\text{ plus or minus } (\frac{3}{4}) x \]
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running along the rectangle's diagonals
\[ y=\pm\tfrac{3}{4}x \]
Verify: check against the curve's behaviour
Why: Far from the centre the equation is close to the difference of the two squared terms being zero, which factors into exactly these two lines. The rectangle construction and the algebra agree.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1223-1225
Prediction
A horizontal hyperbola with a of 5 and b of 2.
Predict first
What are the asymptote slopes?
Correct: Plus and minus two fifths.
Why: The rectangle extends five either side and two up and down, so a corner sits at five across and two up — a slope of two over five. Drawing the rectangle makes the ratio unambiguous.
Worked example
The slope ratio inverts.
\[ \text{Find the asymptotes of } \frac{y^2}{9}-\frac{x^2}{16}=1. \]
Read the constants
Why: Positive term first.
Build the rectangle
Why: Three up and down, four either side.
\[ 8\text{ by } 6 \]
Find the diagonal slopes
Why: Vertical extent over horizontal.
\[ \frac{3}{4} \]
Write the lines
Why: Through the centre.
\[ y =\text{ plus or minus } (\frac{3}{4}) x \]
Figure (svg): The solution to Worked example the other orientation shown as a ladder of expressions, one row per legal move
\[ y=\pm\tfrac{3}{4}x \]
Verify: compare the two hyperbolas
Why: Both share the same central rectangle and therefore the same asymptotes, but they open along perpendicular axes — one pair of branches sits left and right, the other above and below. The rectangle constrains the asymptotes without fixing the orientation.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1225-1226
Error analysis
A student finds the asymptotes of a horizontal hyperbola.
Annotate
On: \( \frac{x^2}{16}-\frac{y^2}{9}=1 \;\Longrightarrow\; y=\pm\tfrac{4}{3}x \)
Recalling the ratio as a formula invites inversion; reading it off a drawn rectangle does not. The corner's coordinates give rise over run directly.
Faded example
With a of 4 horizontally and b of 3 vertically.
Fill in the blanks
\text82a=6, \quad\text___2b=___
Why: The rectangle extends a in each direction along the axis of opening and b in each perpendicular direction, so its full width and height are twice those constants. Its diagonals are the asymptotes.
Sorting
One construction, several features.
Sort into buckets
Sort each item.
Explain it to yourself
An ellipse has none and a hyperbola has two.
Discussion prompt
Explain what makes the difference.
Hint: How far can the curve go?
Answer:
An ellipse is bounded, so it never goes far enough to approach anything. There is nothing for an asymptote to describe.
A hyperbola's branches run off indefinitely, and far from the centre the constant in the equation becomes negligible compared with the squared terms.
What remains is the difference of the two squared terms being zero, which factors into two straight lines. A good explanation notes that the asymptotes are what the equation reduces to at large distances, which is why the branches approach them.
Section
Section 5
Concept
Replacing each variable by its difference from the centre's coordinate translates the whole configuration — branches, vertices, foci and asymptotes together.
Forgetting that the asymptotes move is the commonest translation error. They pass through the centre, not the origin, so a translated hyperbola's asymptotes have nonzero intercepts.
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running along the rectangle's diagonals
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1226-1229
Picture it
Rectangle first, then diagonals, then branches.
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running along the rectangle's diagonals
Translating the picture moves all of it together. Drawing the centre first and building outward is what keeps the asymptotes attached to the right point.
Worked example
Read the centre, then everything else relative to it.
\[ \text{Describe } \frac{(x-1)^2}{9}-\frac{(y+2)^2}{16}=1. \]
Read the centre
Why: Opposite signs to those shown.
\[ (1, -2) \]
Find the positive term
Why: The horizontal one.
Read the constants
Why: Square roots.
\[ a = 3, b = 4 \]
Locate the vertices
Why: Three either side of the centre.
\[ (-2, -2)\text{ and } (4, -2) \]
Figure (svg): A hyperbola with its central rectangle drawn and the asymptotes running along the rectangle's diagonals
\[ \text{centre }(1,-2),\; a=3,\; b=4 \]
Verify: check a vertex
Why: Substituting (4, -2) gives 9 over 9 minus zero, which is 1 — so it is on the curve. The vertices sit three units either side of the centre along the horizontal, as the positive term requires.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1227-1228
Prediction
A hyperbola is centred away from the origin.
Predict first
Through what point do its asymptotes pass?
Correct: The centre.
Why: The asymptotes are the diagonals of the central rectangle, which is built around the centre. Translating the hyperbola moves the rectangle and its diagonals along with it.
Worked example
They pass through the centre, not the origin.
\[ \text{Find the asymptotes of that hyperbola.} \]
Find the slopes
Why: From the constants.
\[ \text{plus and minus } \frac{4}{3} \]
Note they pass through the centre
Why: Not the origin.
\[ \text{through } (1, -2) \]
Write in point-slope form
Why: Using the centre.
\[ y + 2 = \pm(\frac{4}{3}) (x - 1) \]
Check
Why: The centre satisfies both.
Figure (svg): The solution to Worked example translated asymptotes shown as a ladder of expressions, one row per legal move
\[ y+2=\pm\tfrac{4}{3}(x-1) \]
Verify: substitute the centre
Why: At the centre both sides are zero, so both asymptotes pass through it as required. Writing them through the origin instead would displace them from the branches entirely, which is the standard translation error.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1228-1229
Trap
\[ \text{centre }(1,-2), \text{ asymptotes }y=\pm\tfrac{4}{3}x \]
Use the untranslated asymptote equations
Why: The slopes are computed correctly but the lines are placed through the origin.
The asymptotes no longer touch the branches, and the sketch is incoherent.
The asymptotes pass through the centre, so they translate along with everything else.
Writing them in point-slope form from the centre makes this automatic and avoids computing an intercept.
Substituting the centre is the check: both asymptote equations should be satisfied there.
Faded example
Slope four thirds through the centre at one and negative two.
Fill in the blanks
y+2=\tfrac1___(x-___)
Why: Point-slope form from the centre gives the asymptote directly, with the same sign convention as the hyperbola's own equation. Substituting the centre makes both sides zero, which is the check.
Sorting
Positions move and shapes do not.
Sort into buckets
Sort each feature.
Explain it
There is a fastest way to draw a hyperbola.
Discussion prompt
Explain the order to a classmate and why it works.
Hint: What should be drawn before the curve?
Answer:
Centre, rectangle, asymptotes, branches. Mark the centre, extend a and b in the four directions to make the rectangle, draw its diagonals extended, and then sketch the branches.
The rectangle fixes the vertices at the midpoints of two of its sides and the slopes of the asymptotes at once, so the branches have somewhere to start and somewhere to head.
Sketching the branches first leaves both undetermined and usually produces the wrong curvature. A good explanation notes that no points need computing at all — the construction gives an accurate sketch by itself.
Comparison
Fill the blanks from memory. One word of the definition separates them.
Comparison matrix
| ellipse | hyperbola | |
|---|---|---|
| definition | constant sum | constant difference |
| equation sign | plus between the terms | minus between the terms |
| orientation from | the larger denominator | the positive term |
| relation | a squared is the sum | c squared is the sum |
The third row is where habits transfer wrongly. The two conics answer the orientation question by different features, and importing one rule into the other section gives a plausible wrong answer.
Pattern
Five steps, and the second differs from the ellipse's.
Step 4 adds where the ellipse subtracted, and step 2 uses the sign where the ellipse used the size. Those two differences account for most of the errors in this section.
OpenStax Algebra and Trigonometry 2e, §12.2 The Hyperbola §12.2
Check
The definition.
Check your understanding
What is constant for every point on a hyperbola?
Answer: A
Why: A constant difference imposes no bound on either distance, which is why the curve is unbounded and comes in two branches. A constant sum would define an ellipse.
Check
Orientation.
Check your understanding
What determines which way a hyperbola opens?
Answer: A
Why: The positive term names the axis the branches open along, regardless of the denominators' relative sizes. Using the larger denominator is the ellipse's rule and gives the wrong axis whenever the sizes happen to be ordered the other way.
Check
The relation.
Check your understanding
For a hyperbola, how is the focal distance found?
Answer: A
Why: The hyperbola's relation puts the focal distance alone as the hypotenuse, so its square is the sum. That makes it larger than the semi-transverse axis, which is why each focus lies beyond its vertex.
Real world
Long-range navigation systems located ships by hyperbolas.
Discussion prompt
Two transmitters send synchronised pulses and a receiver measures the difference in arrival times. What does that determine?
Hint: A time difference is a distance difference.
Answer:
A difference in arrival times is a difference in distances, since the pulses travel at a known speed. So the receiver lies on a curve of constant distance difference — a hyperbola with the two transmitters as foci.
One measurement gives a whole hyperbola rather than a point. A second pair of transmitters gives another hyperbola, and the intersection fixes the position.
This was the basis of the Loran systems used at sea for decades. The focal definition is doing the work directly, and modern satellite positioning uses the same principle with spheres instead of hyperbolas.
Commit first
State your confidence along with your answer.
Predict first
Why does a hyperbola have asymptotes when an ellipse does not?
Correct: Its branches are unbounded, so they approach limiting directions.
Why: Far from the centre the constant becomes negligible and the equation reduces to the two squared terms being equal, which factors into two straight lines. An ellipse is bounded and never travels far enough for anything to approach.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate used the larger denominator to decide which way a hyperbola opens. Explain the mistake.
Hint: Which rule belongs to which conic?
Answer:
That is the ellipse's rule, where both terms are positive so only the sizes distinguish them.
For a hyperbola one term is negative, and the positive one names the axis of opening. The denominators may be in any size order at all.
So the two conics answer the same question by different features. A good explanation suggests naming the conic first — plus sign or minus sign between the terms — since that determines which rule applies before anything else is read.
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 second and third are where habits from §10.1 transfer wrongly, so they cost the most. The fourth is the technique that makes an accurate sketch possible without computing points.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a hyperbola with its centre, vertices, foci, central rectangle and asymptotes marked. Beside it, write a two-column comparison with the ellipse covering the definition, the orientation rule and the relation among the constants. Underneath, sketch one translated hyperbola in the order centre, rectangle, asymptotes, branches.
If your comparison names the orientation rule as sign for one conic and size for the other, the section's most transferable error is accounted for.
Recap
Five things, and two of them differ from the ellipse in ways worth marking.
| if you remember one thing | it should be this |
|---|---|
| about the definition | constant difference, so the curve is unbounded |
| about orientation | the positive term names the axis; size is irrelevant |
| about the relation | the focal distance is the hypotenuse, so add |
| about sketching | rectangle first, then diagonals, then branches |
Section 10.3 completes the three conics with the parabola, whose definition uses one focus and a line rather than two foci — and which therefore has no centre and no asymptotes at all.
OpenStax, Precalculus, §10.2 The Hyperbola §10.2, pp. 1208-1229 — everything on these slides traces back here
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