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Ellipse: definition, axes and constraints

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

高二選擇性必修第一册(A版).pdf · 3.1 · PDF 110 / printed page 105

TOPIC 01

Ellipse: definition, axes and constraints

Build an ellipse from a distance sum, identify its major direction and distinguish a genuine ellipse from boundary loci.

What you will be able to explain

Relate the focal definition to a²=b²+c².
Read translated and vertical ellipses without confusing denominators with lengths.
Use scaling to optimize a linear quantity and test locus feasibility.

A fixed sum is possible only above the focal separation

The definition lets us reject impossible data before manipulating two square roots.

For foci (±c,0), a nondegenerate ellipse has distance sum 2a>2c. Its transverse semiaxis satisfies b²=a²−c².

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

At a major vertex the focal distances add to 2a.
$(a-c)+(a+c)=2a$
At a minor vertex each focal distance is a.
$b^2+c^2=a^2$

When it applies

Use a for the larger semiaxis when naming focal quantities.
Equal semiaxes give a circle; focal sum equal to separation gives a segment.

Counterexample: For foci separated by 6, a distance sum of 5 is impossible and a sum of 6 gives a segment, not a nondegenerate ellipse.

Distance-sum definition ↔ standard equation ↔ scaled unit circle.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict which axis contains the foci when the two semiaxis controls exchange size.

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

The major axis follows the larger semiaxis. A zero semiaxis is a boundary segment (both zero gives a point); equal positive axes give a circle.

Numerical data for the diagram; printable
Quantity	Value
a_horizontal / 水平半軸	4
b_vertical / 垂直半軸	2
c	3.46410162
PF1+PF2	8
2 × major semiaxis / 長半軸兩倍	8

Explain: The focal separation is computed from the difference of squared semiaxes; a zero semiaxis is shown as a boundary segment.

Transfer: Find the circular setting, then explain why its coincident foci still satisfy the distance-sum definition.

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

Identify the major axis, foci and eccentricity.

\frac{x^2}{25}+\frac{y^2}{9}=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Compare the denominators.

Hint 2

The focal square is their positive difference.

Worked solution

The major semiaxis is 5 along x; the minor is 3.
$c^2=25-9=16$
Locate the symmetric foci and divide c by a.
$F=(\pm4,0),\ e=4/5$

Horizontal major axis; foci (±4,0); e=4/5.

Checks and common pitfalls: The denominator is a squared semiaxis, not a full axis length.

Reasoning checklist · self / teacher assessment

Horizontal major axis; foci (±4,0); e=4/5.
The denominator is a squared semiaxis, not a full axis length.

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

02 / Standard#Worked example

An ellipse has foci (±3,0) and distance sum 10. Derive its equation.

First step provided — complete the reasoning:Read a=5 and c=3.

b^2=25-9=16

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The distance sum equals 2a.

Hint 2

Use b²=a²−c².

Worked solution

Read a=5 and c=3.
$b^2=25-9=16$
Use the horizontal-major-axis form.
$x^2/25+y^2/16=1$

x²/25+y²/16=1.

Checks and common pitfalls: Using a=10 would double the intended distance sum.

Reasoning checklist · self / teacher assessment

x²/25+y²/16=1.
Using a=10 would double the intended distance sum.

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

03 / Transfer#Worked example

Find the greatest possible distance from a point on this ellipse to its right focus.

x^2/25+y^2/9=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The farthest vertex is on the opposite end of the major axis.

Hint 2

Use the focal-distance relation r=a−ex.

Worked solution

Here a=5,c=4 and −5≤x≤5.
$r=5-4x/5$
The minimum x gives the maximum r.
$r_{\max}=5+4=9$

Maximum 9 at (−5,0).

Checks and common pitfalls: The focal sum 10 is not either individual focal distance.

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

04 / Foundation#Your turn

Find c, the centre-to-focus distance.

x^2/16+y^2/7=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Subtract the squared semiaxes.

Hint 2

c is positive.

Worked solution

Use the ellipse relation.
$c^2=16-7=9$
Take the positive root.
$c=3$

c=3.

Checks and common pitfalls: The plus relation belongs to a hyperbola.

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

05 / Foundation#Your turn

Find the foci of the vertical ellipse.

x^2/9+y^2/25=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The larger denominator is under y².

Hint 2

Place c on the vertical axis.

Worked solution

Calculate c².
$c^2=25-9=16$
Place the foci about the origin.
$F_1=(0,-4),\ F_2=(0,4)$

(0,±4).

Checks and common pitfalls: The letter x does not automatically indicate the major axis.

Reasoning checklist · self / teacher assessment

(0,±4).
The letter x does not automatically indicate the major axis.

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

06 / Foundation#Your turn

Find the constant sum of focal distances.

x^2/36+y^2/20=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Read the major semiaxis.

Hint 2

The defining sum is twice it.

Worked solution

The major semiaxis is √36.
$a=6$
Use the focal definition.
$r_1+r_2=2a=12$

12.

Checks and common pitfalls: The sum is independent of the point on the ellipse.

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

07 / Foundation#Your turn

The point (3,y) lies on the ellipse. Find y².

x^2/25+y^2/16=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute x=3.

Hint 2

Solve for y² before choosing a sign.

Worked solution

Rearrange the residual fraction.
$y^2/16=1-9/25=16/25$
Multiply by the denominator.
$y^2=256/25$

y²=256/25.

Checks and common pitfalls: Both signs of y are possible; its square is unique.

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

08 / Standard#Your turn

For which k is the locus a noncircular, nondegenerate ellipse?

x^2/9+y^2/k=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Both denominators must be positive.

Hint 2

Equal denominators give a circle.

Worked solution

Exclude zero and negative k.
$k>0$
Exclude the circle while permitting either major-axis direction.
$k\ne9$

k>0 and k≠9.

Checks and common pitfalls: k>9 is also an ellipse; it simply has a vertical major axis.

Reasoning checklist · self / teacher assessment

k>0 and k≠9.
k>9 is also an ellipse; it simply has a vertical major axis.

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

09 / Standard#Your turn

Find the eccentricity.

x^2/100+y^2/36=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Find c before dividing.

Hint 2

Use c/a, not b/a.

Worked solution

Find the focal distance.
$c=\sqrt{100-36}=8$
Divide by the major semiaxis.
$e=8/10=4/5$

e=4/5.

Checks and common pitfalls: For a noncircular ellipse, 0<e<1.

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

10 / Standard#Your turn

Find the centre, major-axis vertices and foci.

(x-2)^2/16+(y+1)^2/7=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Subtract the centre shift before reading parameters.

Hint 2

The focal distance is 3.

Worked solution

The centre is (2,−1), with a=4.
$V=(2\pm4,-1)$
Translate both foci consistently.
$F=(2\pm3,-1)$

Centre (2,−1); major-axis vertices (−2,−1),(6,−1); foci (−1,−1),(5,−1).

Checks and common pitfalls: The sign in y+1 places the centre at y=−1.

Reasoning checklist · self / teacher assessment

Centre (2,−1); major-axis vertices (−2,−1),(6,−1); foci (−1,−1),(5,−1).
The sign in y+1 places the centre at y=−1.

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

11 / Standard#Your turn

The foci are (−3,0),(3,0), and the focal-distance sum is 6. Is the locus a nondegenerate ellipse? Describe it.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The sum equals the distance between foci.

Hint 2

Use the equality case of the triangle inequality.

Worked solution

Equality forces the point onto the segment joining the foci.
$r_1+r_2=|F_1F_2|$
Every point of that closed segment has the required sum.

The closed segment −3≤x≤3,y=0, not a nondegenerate ellipse.

Checks and common pitfalls: A genuine ellipse needs the distance sum strictly larger than the focal separation.

Reasoning checklist · self / teacher assessment

The closed segment −3≤x≤3,y=0, not a nondegenerate ellipse.
A genuine ellipse needs the distance sum strictly larger than the focal separation.

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

12 / Transfer#Your turn

Find the maximum of 3x+4y on the ellipse, and give an attaining point.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Set x=2u,y=3v with u²+v²=1.

Hint 2

Apply the dot-product bound in scaled coordinates.

Worked solution

Set u=x/2 and v=y/3, so u²+v²=1; the objective becomes a dot product with (6,12).
$3x+4y=6u+12v\le\sqrt{180}=6\sqrt5$
Equality requires parallel unit directions.
$(u,v)=(1/\sqrt5,2/\sqrt5)$

Maximum 6√5 at (2/√5,6/√5).

Checks and common pitfalls: Bounding 3x and 4y separately gives an unattainable corner.

Reasoning checklist · self / teacher assessment

Maximum 6√5 at (2/√5,6/√5).
Bounding 3x and 4y separately gives an unattainable corner.

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

13 / Transfer#Your turn

P has distance sum 10 to (±3,0) and lies above the x-axis at x=0. Find P, then explain why replacing sum 10 by 5 makes the locus empty.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

At x=0 both distances are equal.

Hint 2

Compare the proposed sum 5 with focal separation 6.

Worked solution

Each distance is 5, so y²+9=25.
$P=(0,4)$
Any two focal distances sum to at least the focal separation.
$r_1+r_2\ge6>5$

P=(0,4); the sum-5 locus is empty.

Checks and common pitfalls: Writing a standard-looking equation does not guarantee a geometrically possible locus.

Reasoning checklist · self / teacher assessment

P=(0,4); the sum-5 locus is empty.
Writing a standard-looking equation does not guarantee a geometrically possible locus.

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	Predict the string-and-pins boundary cases.
8	Concept derivation	Derive the relation between a,b,c from axis vertices.
10	Guided application	Read horizontal, vertical and translated equations.
12	Transfer and explanation	Solve a scaled optimization and an impossible-locus test.
5	Exit evidence	Classify the focal-sum exit cases.

Question sequence

What happens when the string length equals the distance between the pins?
Which direction has the largest intercept?

Board development

Use three rows: sum below, equal to, and above the focal separation.
Scale x=2U,y=3V before bounding 3x+4y.

Expected correct and incorrect responses

Students always put foci on the x-axis. Ask them to mark the larger intercept first.
Correct: “The larger denominator is under y², so the foci lie on the y-axis.”
Incorrect: “The denominator 25 means the semiaxis has length 25.”

Follow-up questions

Why must an equality case be checked after finding an upper bound?

Remediation

Draw semiaxes as actual lengths before calculating c².

Exit check

Classify distance sums 5, 6 and 10 for foci at (±3,0).

Assessment checklist and sample responses

Credit axis identification, positive-length conditions and an attainable equality case. Sample: 6U+12V≤√180 with U²+V²=1; equality fixes the maximizing point.
Correct: “The larger denominator is under y², so the foci lie on the y-axis.”