LEARN · EXPLAIN · REVISE

All lines through P=(0,1) cut the ellipse in a chord. Find the locus of its midpoint M=(h,k), including the vertical line case.

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

高二選擇性必修第一册(A版).pdf · 3.1 · PDF 110 / printed page 105

Revisit first: Ellipse: definition, axes and constraints Hyperbola: branches and asymptotes Parabola: focus and directrix

TOPIC 01

Conic intersections, chords and inverse problems

Solve line–conic systems with degree and domain checks; use Vieta for midpoint loci and reconstruct a chord from its midpoint.

What you will be able to explain

Retain all roots and substitute them into the original system.
Use the discriminant only for an actual quadratic.
Use root sums and products without unnecessarily solving every endpoint.

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.

01 / Transfer#Your turn

All lines through P=(0,1) cut the ellipse in a chord. Find the locus of its midpoint M=(h,k), including the vertical line case.

x^2/9+y^2/4=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

For y=mx+1 use Vieta to obtain h,k.

Hint 2

Eliminate m, then restore any missing vertical case.

Worked solution

The substituted quadratic gives the midpoint parametrization.
$h=-9m/(4+9m^2),\ k=4/(4+9m^2)$
Eliminate m and complete a square.
$h^2/9+k(k-1)/4=0$
Finite m gives k>0; the vertical chord adds (0,0). All other locus points recover m=−4h/(9k).
$h^2/(9/16)+(k-1/2)^2/(1/4)=1$

The complete ellipse h²/(9/16)+(k−1/2)²/(1/4)=1.

Checks and common pitfalls: Elimination alone can hide a missing point caused by slope form.

Reasoning checklist · self / teacher assessment

The complete ellipse h²/(9/16)+(k−1/2)²/(1/4)=1.
Elimination alone can hide a missing point caused by slope form.

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	Find the lost zero root in a proposed solution.
8	Concept derivation	Derive the ellipse-line quadratic and Vieta midpoint.
10	Guided application	Classify secant, tangent, external and degree-loss cases.
12	Transfer and explanation	Reconstruct a chord and prove the full midpoint locus including the vertical case.
5	Exit evidence	Write the degree condition beside the discriminant rule.

Question sequence

What is the actual degree after substitution?
Can the midpoint be found without writing both radicals?

Board development

Place degree check before discriminant, then back-substitution after roots.
For the midpoint locus, reserve a box for the vertical chord and the reverse parametrization.

Expected correct and incorrect responses

Students prove only that chord midpoints satisfy an ellipse equation. Ask how an arbitrary point on that ellipse recovers a chord.
Correct: “I keep x=0 when factoring x(x−4)=0, and then use y=x.”
Incorrect: “Dividing by x gives x=4, so (4,4) is the only intersection.”

Follow-up questions

Which datum determines the slope in the inverse-parabola chord problem?

Remediation

Factor x(x−4)=0 before any division; check each point in y=x.

Exit check

State the extra check needed before using Δ=0 as a tangency test.

Assessment checklist and sample responses

Credit degree, full root retention, back-substitution and a two-way locus argument. Sample: finite m covers k>0; the vertical chord supplies (0,0), completing the midpoint locus.
Correct: “I keep x=0 when factoring x(x−4)=0, and then use y=x.”