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After line substitution the equation is 0x²+2x−4=0. How many real intersections follow?

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

Revisit first: Ellipse: definition, axes and constraintsHyperbola: branches and asymptotesParabola: 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.

01 / Foundation#Your turn

After line substitution the equation is 0x²+2x−4=0. How many real intersections follow?

Skills and prerequisite lessons

Scope: JM01 / JM02

  1. Option A00
  2. Option B11
  3. Option C22
  4. Option D∞\infty
Working and explanation
BUILD THE REASONING
Hint 1
Identify the actual degree first.
Hint 2
Solve 2x−4=0.
Worked solution
  1. The equation is linear.
    x=2x=2
  2. A line gives one corresponding ordinate.
B: one intersection; tangency does not follow from this alone.
Checks and common pitfalls: The discriminant criterion requires a nonzero quadratic coefficient.
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 diagnosisFind the lost zero root in a proposed solution.
8Concept derivationDerive the ellipse-line quadratic and Vieta midpoint.
10Guided applicationClassify secant, tangent, external and degree-loss cases.
12Transfer and explanationReconstruct a chord and prove the full midpoint locus including the vertical case.
5Exit evidenceWrite 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.”
No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.
Curriculum and source notes ↗