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Choose the vertex of (y−2)²=4(x+3).

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 · 3.3 · PDF 135 / printed page 130

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.”