LEARN · EXPLAIN · REVISE

Hyperbola: branches and asymptotes

Read the idea, work independently, then explain what changed.

高二選擇性必修第一册(A版).pdf · 3.2 · PDF 123 / printed page 118

Revisit first: Ellipse: definition, axes and constraints

TOPIC 01

Hyperbola: branches and asymptotes

Keep signed and absolute focal differences separate; inspect degree loss when a line is parallel to an asymptote.

What you will be able to explain

Derive c²=a²+b² from the focal difference.
Identify branches, vertices and asymptotes in both orientations.
Classify focal-difference boundary cases and one-intersection lines.

Difference fixes two branches; sign selects one

A line may meet a hyperbola once because a quadratic term cancels, so a one-point count alone cannot certify tangency.

A nondegenerate hyperbola requires 0<2a<2c. An absolute difference gives two branches; assigning a sign distinguishes which focus is farther.

x^2/a^2-y^2/b^2=1,\quad c^2=a^2+b^2,\quad y=\pm(b/a)x

Rearrange the equation to compare large coordinate magnitudes.
$y^2=(b^2/a^2)x^2-b^2$
The leading ratio approaches ±b/a, but substituting the asymptote exactly produces 0=1.
$x^2/a^2-(bx/a)^2/b^2=0$

When it applies

Both a and b must be positive for the displayed standard equation.
Check the actual polynomial degree after line substitution.

Counterexample: The line y=x+1 meets x²−y²=1 only at (−1,0), but the tangent there is x=−1.

Focal difference ↔ signed branch ↔ standard equation ↔ asymptote slopes.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict whether shifting an asymptote parallel to itself produces zero, one or two intersections.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

The red line is parallel to the positive-slope asymptote. At zero offset it has no intersection; otherwise the cancelled quadratic leaves one linear solution, not a tangency test.

Numerical data for the diagram; printable
Quantity	Value
slope / 斜率	0.5
offset / 截距	1
intersection count / 交點數	1
P	-2, 0

Explain: Exact cancellation leaves a linear equation for nonzero offset and a contradiction at zero offset.

Transfer: Explain why the intersection escapes the viewing window as the offset approaches zero.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

Find the vertices, foci and asymptotes.

x^2/9-y^2/16=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The positive term sets the transverse axis.

Hint 2

For a hyperbola, c² is a sum.

Worked solution

Read a=3,b=4,c=5.
$c^2=9+16=25$
Place the vertices/foci on x and set the homogeneous quadratic to zero.
$V=(\pm3,0),\ F=(\pm5,0),\ y=\pm4x/3$

Vertices (±3,0), foci (±5,0), asymptotes y=±4x/3.

Checks and common pitfalls: The asymptotes pass through the centre, not a focus.

Reasoning checklist · self / teacher assessment

Vertices (±3,0), foci (±5,0), asymptotes y=±4x/3.
The asymptotes pass through the centre, not a focus.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

The foci are (±5,0), and the absolute focal-distance difference is 6. Find the equation.

First step provided — complete the reasoning:Use a=3,c=5.

b^2=25-9=16

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The constant is 2a.

Hint 2

Recover b²=c²−a².

Worked solution

Use a=3,c=5.
$b^2=25-9=16$
Keep both branches because the difference is absolute.
$x^2/9-y^2/16=1$

x²/9−y²/16=1.

Checks and common pitfalls: A signed distance difference would select one branch.

Reasoning checklist · self / teacher assessment

x²/9−y²/16=1.
A signed distance difference would select one branch.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

The line y=x+1 meets x²−y²=1 once. Is it tangent? Justify.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitution may reduce the degree.

Hint 2

Compare the actual tangent slope at the intersection.

Worked solution

The quadratic terms cancel, leaving one root.
$x^2-(x+1)^2=1\Rightarrow x=-1,\quad y=0$
The tangent at (−1,0) is vertical, whereas y=x+1 is not.
$x=-1$

Not tangent; it is parallel to an asymptote.

Checks and common pitfalls: One intersection does not imply tangency when the substituted equation becomes linear.

Reasoning checklist · self / teacher assessment

Not tangent; it is parallel to an asymptote.
One intersection does not imply tangency when the substituted equation becomes linear.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the centre-to-focus distance c.

x^2/16-y^2/9=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Add the squared parameters.

Hint 2

Take the positive square root.

Worked solution

Use c²=a²+b².
$c^2=16+9=25$
Convert to distance.
$c=5$

c=5.

Checks and common pitfalls: Subtracting would use the ellipse relation.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Find the eccentricity.

x^2/9-y^2/16=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The focus distance is 5.

Hint 2

Divide by the transverse semiaxis 3.

Worked solution

Find c/a.
$e=\sqrt{9+16}/3$
Simplify.
$e=5/3$

e=5/3.

Checks and common pitfalls: A hyperbola has e>1, so 3/5 cannot be correct.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Find the vertices and asymptotes.

y^2/4-x^2/9=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The positive term is vertical.

Hint 2

Set the quadratic difference to zero for asymptotes.

Worked solution

The vertices have x=0.
$V=(0,\pm2)$
Solve the homogeneous equation.
$y^2/4=x^2/9\Rightarrow y=\pm2x/3$

Vertices (0,±2); asymptotes y=±2x/3.

Checks and common pitfalls: Do not interchange the slope ratio merely because the major letter changes.

Reasoning checklist · self / teacher assessment

Vertices (0,±2); asymptotes y=±2x/3.
Do not interchange the slope ratio merely because the major letter changes.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

A point (5,y) lies on x²/9−y²/16=1. Find y².

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Move the negative term to the other side.

Hint 2

Keep exact fractions.

Worked solution

Substitute x=5.
$y^2/16=25/9-1=16/9$
Multiply by 16.
$y^2=256/9$

y²=256/9.

Checks and common pitfalls: There are two y values; neither changes the branch x>0.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Describe the possible x-coordinates on the hyperbola.

x^2/9-y^2/16=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The y² term is nonnegative.

Hint 2

Solve the resulting square inequality as a union.

Worked solution

Rearrange to bound x².
$x^2/9=1+y^2/16\ge1$
Every x in the resulting rays gives real y.
$x\le-3\text{ or }x\ge3$

(−∞,−3]∪[3,∞).

Checks and common pitfalls: Using −3≤x≤3 would describe the excluded central strip.

Reasoning checklist · self / teacher assessment

(−∞,−3]∪[3,∞).
Using −3≤x≤3 would describe the excluded central strip.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Prove that no finite point of y=4x/3 lies on this hyperbola.

x^2/9-y^2/16=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute the line exactly.

Hint 2

The left side becomes a constant.

Worked solution

Both squared terms become equal.
$x^2/9-(4x/3)^2/16=0$
The equation would require 0=1.

There is no finite intersection.

Checks and common pitfalls: An asymptote is approached at infinity; it is not part of the curve.

Reasoning checklist · self / teacher assessment

There is no finite intersection.
An asymptote is approached at infinity; it is not part of the curve.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

A horizontal hyperbola has a=3 and perpendicular asymptotes. Find b.

a>0,\ b>0

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The slopes are b/a and −b/a.

Hint 2

Their product is −1.

Worked solution

Use perpendicular slopes.
$-b^2/9=-1$
Choose the positive parameter.
$b=3$

b=3.

Checks and common pitfalls: The negative root does not represent a positive semiaxis parameter.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Which branch satisfies PF₁−PF₂=6?

F_1=(-5,0),F_2=(5,0),\ x^2/9-y^2/16=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Test the right vertex.

Hint 2

The signed difference is constant on each branch.

Worked solution

At (3,0), the distances are 8 and 2.
$PF_1-PF_2=6$
The right branch is farther from the left focus.
$x\ge3$

The right branch, x≥3.

Checks and common pitfalls: The absolute-value equation alone includes both branches.

Reasoning checklist · self / teacher assessment

The right branch, x≥3.
The absolute-value equation alone includes both branches.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

A centred horizontal hyperbola has asymptotes y=±2x and passes through (√2,2). Find its equation.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The slopes give b=2a.

Hint 2

The point fixes the remaining scale.

Worked solution

Insert b²=4a² into the point equation.
$2/a^2-4/(4a^2)=1$
Recover both parameters.
$a^2=1,\ b^2=4$

x²−y²/4=1.

Checks and common pitfalls: Asymptotes determine shape but not scale; a point is also needed.

Reasoning checklist · self / teacher assessment

x²−y²/4=1.
Asymptotes determine shape but not scale; a point is also needed.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Two foci are 10 units apart. Classify loci with absolute distance difference d for d=0,6,10,12.

F_1=(-5,0),\ F_2=(5,0)

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The reverse triangle inequality bounds d by 10.

Hint 2

Analyze both equality boundaries geometrically.

Worked solution

Zero difference gives the perpendicular bisector; an interior positive difference gives a hyperbola.
$d=0:x=0;\ d=6:x^2/9-y^2/16=1$
Maximum difference occurs on the exterior focal rays; larger differences are impossible.
$d=10:y=0,|x|\ge5;\ d=12:\varnothing$

Line; nondegenerate hyperbola; two exterior rays; empty set, respectively.

Checks and common pitfalls: The defining constant must satisfy 0<d<10 for a nondegenerate hyperbola.

Reasoning checklist · self / teacher assessment

Line; nondegenerate hyperbola; two exterior rays; empty set, respectively.
The defining constant must satisfy 0<d<10 for a nondegenerate hyperbola.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Prediction and diagnosis	Locate the sign of the focal difference at each vertex.
8	Concept derivation	Derive and sketch asymptote slopes with horizontal and vertical axes.
10	Guided application	Calculate a point and test whether an asymptote intersects.
12	Transfer and explanation	Classify the offset-line and focal-difference boundary cases.
5	Exit evidence	Explain why a single intersection is insufficient for tangency.

Question sequence

Which focus is farther on the right branch?
Can a line have one intersection without being tangent?

Board development

Tabulate absolute difference 0, between 0 and 2c, equal to 2c and greater than 2c.
Substitute y=x+k before mentioning any discriminant.

Expected correct and incorrect responses

Students set the discriminant to zero even after the x² term disappears. Circle the actual leading coefficient first.
Correct: “The quadratic coefficient cancels, so one intersection need not be tangency.”
Incorrect: “A line with exactly one intersection must be tangent.”

Follow-up questions

How does the unique intersection move as the parallel line offset tends to zero?

Remediation

Use the vertices to decide which signed focal difference is positive.

Exit check

Give an equation of a one-intersection non-tangent line and explain the degree loss.

Assessment checklist and sample responses

Require branch sign, standard-form parameters and a separate tangency justification. Sample: substituting y=x+1 gives −2x−1=1, so x=−1; this linear equation alone does not imply tangency.
Correct: “The quadratic coefficient cancels, so one intersection need not be tangency.”