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Choose the major semiaxis length for x²/9+y²/25=1.

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

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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.

Try it. Leave your reasoning visible.

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

01 / Foundation#Your turn

Choose the major semiaxis length for x²/9+y²/25=1.

Skills and prerequisite lessons

Scope: JM01 / JM02

  1. Option A33
  2. Option B55
  3. Option C99
  4. Option D2525
Working and explanation
BUILD THE REASONING
Hint 1
The larger denominator identifies the major axis.
Hint 2
Take its square root.
Worked solution
  1. Compare 9 and 25.
    a2=25a^2=25
  2. Convert squared length to length.
    a=5a=5
B: 5.
Checks and common pitfalls: A full major axis would have length 10.
Think first. Reveal a hint when the class is ready.
Focus on one question
Teacher preparation and assessment
40-minute lesson
MinutesFocusTeaching and student action
5Prediction and diagnosisPredict the string-and-pins boundary cases.
8Concept derivationDerive the relation between a,b,c from axis vertices.
10Guided applicationRead horizontal, vertical and translated equations.
12Transfer and explanationSolve a scaled optimization and an impossible-locus test.
5Exit evidenceClassify 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.”
No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.
Curriculum and source notes ↗