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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.
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.
Checks and common pitfalls: The sum is independent of the point on the ellipse.
Think first. Reveal a hint when the class is ready.
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.”