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Choose c² for x²/25+y²/9=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 c² for x²/25+y²/9=1.

Skills and prerequisite lessons

Scope: JM01 / JM02

  1. Option A3434
  2. Option B1616
  3. Option C44
  4. Option D2525

Working and explanation

BUILD THE REASONING

Hint 1
This is an ellipse with positive squared terms.
Hint 2
Subtract minor from major squared semiaxis.
Worked solution
  1. Apply the focal relation.

    c2=a2−b2c^2=a^2-b^2
  2. Calculate the difference.

    25−9=1625-9=16

B: 16.

Checks and common pitfalls: The focus lies inside the major vertex.

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 ↗