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TOPIC 01
Parabola: focus and directrix
Derive a parabola from equal distances, track orientation under translation and use the definition in a distance optimization.
What you will be able to explain
Convert between y²=4ax and its focus/directrix.
Retain the absolute distance to a line when deriving a translated equation.
Check attainability in a focal-distance inequality.
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: A negative opening coefficient is valid, not an empty locus.
Reasoning checklist · self / teacher assessment
Focus (−2,0), directrix x=2.
A negative opening coefficient is valid, not an empty locus.
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
Diagnose the factor-of-four focus error.
8
Concept derivation
Derive the focus/directrix equation including absolute distance.
10
Guided application
Read rotated and translated parabolas and their focal chords.
12
Transfer and explanation
Solve the distance minimization and analyze its equality set.
5
Exit evidence
Identify the a=0 boundary locus.
Question sequence
Why is distance to x=−a written with an absolute value?
What changes if the focus is left of the vertex?
Board development
Label vertex, focus, directrix and focal chord on the same diagram.
Keep a separate line for the equality case in the distance optimization.
Expected correct and incorrect responses
Students halve 8 instead of dividing it by 4 to locate the focus. Substitute the proposed focus into the defining geometry.
Correct: “4a=8, so the focus is (2,0) and directrix x=−2.”
Incorrect: “The focus is (8,0) because the equation contains 8x.”
Follow-up questions
What set remains when a=0, and why is it excluded from the nondegenerate definition?
Remediation
Compare the vertex-to-focus distance and the vertex-to-directrix distance before expanding squares.
Exit check
For y²=−8x, state both focus and directrix with their signs.
Assessment checklist and sample responses
Credit parameter interpretation, valid squaring and attained equality. Sample: on y²=8x, PF=x+2; the proposed bound is reached at the vertex and nowhere else.
Correct: “4a=8, so the focus is (2,0) and directrix x=−2.”
