LEARN · EXPLAIN · REVISE

Conic intersections, chords and inverse problems

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.

Elimination preserves a system only when its operations are reversible

A midpoint question usually needs the sum of roots, while an inverse chord question may need both sum and product.

After substituting a line, inspect the degree before counting roots. For a genuine quadratic, Vieta provides the midpoint; a slope parametrization must restore vertical lines separately.

Ax^2+Bx+C=0,\quad x_1+x_2=-B/A\quad(A\ne0)

Substitute y=mx+b in the ellipse equation.
$(4+9m^2)x^2+18mbx+9b^2-36=0$
For two intersections, average the roots without extracting them individually.
$x_M=-9mb/(4+9m^2)$

When it applies

Do not divide by a variable without checking its zero case.
Tangency via a discriminant assumes a nonzero quadratic coefficient.
A locus derivation needs both necessity and sufficiency, including omitted parameter cases.

Counterexample: Dividing x²=4x by x loses the intersection at the origin.

System ↔ eliminated equation ↔ root relations ↔ chord geometry.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict intersection count as the line intercept passes its tangent values.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

The actual quadratic always has positive leading coefficient for this ellipse. Zero discriminant merges the two endpoints; vertical lines are handled separately.

Numerical data for the diagram; printable
Quantity	Value
A	6.25
B	9
C	-27
discriminant / 判別式	756
intersections / 交點數	2
midpoint / 中點	-0.72, 0.64

Explain: The model reports the actual quadratic, discriminant and midpoint, and offers a separate vertical-line mode.

Transfer: Choose a secant and move it to tangency; explain why the two endpoints and their midpoint coincide at the boundary.

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 / Foundation#Worked example

Find both intersections of the ellipse and the line.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute the fixed y-coordinate.

Hint 2

Recover both square-root signs.

Worked solution

The remaining quadratic is in x only.
$x^2/9=3/4$
Pair each x with the original y.
$(x,y)=(\pm3\sqrt3/2,1)$

(±3√3/2,1).

Checks and common pitfalls: An x-root alone is not a complete intersection point.

Reasoning checklist · self / teacher assessment

(±3√3/2,1).
An x-root alone is not a complete intersection point.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Find the intersections, retaining the vertex.

y^2=4x,\ y=x

First step provided — complete the reasoning:The equation includes a zero root.

x^2-4x=x(x-4)=0

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute y=x.

Hint 2

Factor without dividing by x.

Worked solution

The equation includes a zero root.
$x^2-4x=x(x-4)=0$
Return each root to y=x.
$(x,y)=(0,0),(4,4)$

(0,0) and (4,4).

Checks and common pitfalls: Dividing by x would lose the vertex.

Reasoning checklist · self / teacher assessment

(0,0) and (4,4).
Dividing by x would lose the vertex.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

Classify intersections with y=x+k for all real k. Explain why counting one point is insufficient for tangency.

x^2-y^2=1

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Keep the leading coefficient after substitution.

Hint 2

Separate k=0 before dividing.

Worked solution

The quadratic cancels.
$-2kx-k^2=1$
Zero k is inconsistent; nonzero k gives one finite point.
$x=-(1+k^2)/(2k),\ y=(k^2-1)/(2k)$
Slope one cannot be a tangent direction because that would require x=y on this curve.

No intersection for k=0; one nontangent intersection for k≠0.

Checks and common pitfalls: A discriminant test presupposes an actual quadratic.

Reasoning checklist · self / teacher assessment

No intersection for k=0; one nontangent intersection for k≠0.
A discriminant test presupposes an actual quadratic.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the chord length cut by y=0.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The endpoints are on the major axis.

Hint 2

Subtract their coordinates.

Worked solution

Set y=0 to obtain x=±3.
$x^2=9$
Measure between both vertices.
$L=3-(-3)=6$

Length 6.

Checks and common pitfalls: One semiaxis is only half this chord.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

How many real intersections are there?

x^2/9+y^2/4=1,\ y=3

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The y-term already exceeds one.

Hint 2

A real square cannot be negative.

Worked solution

Substitute y=3.
$x^2/9=1-9/4=-5/4$
No real x satisfies it.

Zero intersections.

Checks and common pitfalls: Do not take a real square root of a negative number.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Show that y=x+1 is tangent to y²=4x and find the contact point.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute and collect a genuine quadratic.

Hint 2

A repeated root identifies the contact point here.

Worked solution

The substituted equation is a perfect square.
$(x+1)^2=4x\Rightarrow(x-1)^2=0$
Back-substitute and check the slope.
$(x,y)=(1,2),\ 2y y\prime=4\Rightarrow y\prime=1$

Tangent at (1,2).

Checks and common pitfalls: One must check that the quadratic leading coefficient is nonzero.

Reasoning checklist · self / teacher assessment

Tangent at (1,2).
One must check that the quadratic leading coefficient is nonzero.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Explain why the midpoint of any chord through the centre of an ellipse is the centre.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The equation is unchanged under (x,y)→(−x,−y).

Hint 2

A line through the origin also retains that symmetry.

Worked solution

The two endpoints occur as opposite pairs.
$B=-A$
Their average is the centre.
$(A+B)/2=(0,0)$

The midpoint is (0,0).

Checks and common pitfalls: This argument requires a central chord, not an arbitrary chord.

Reasoning checklist · self / teacher assessment

The midpoint is (0,0).
This argument requires a central chord, not an arbitrary chord.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

For tangency with y=x+b, find b².

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Substitute and keep the parameter b.

Hint 2

Set the quadratic discriminant to zero.

Worked solution

Clear denominators.
$13x^2+18bx+9b^2-36=0$
Solve the zero-discriminant equation.
$\Delta=144(13-b^2)=0$

b²=13, giving b=±√13.

Checks and common pitfalls: Two parallel tangents exist, one on each side.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Find the midpoint of the chord without solving both endpoints.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use the sum of quadratic roots.

Hint 2

The line relates the average y to average x.

Worked solution

Substitution gives root sum −18/13.
$13x^2+18x-27=0$
Halve the sum and use y=x+1.
$M=(-9/13,4/13)$

M=(−9/13,4/13).

Checks and common pitfalls: The midpoint is generally inside the ellipse, not on it.

Reasoning checklist · self / teacher assessment

M=(−9/13,4/13).
The midpoint is generally inside the ellipse, not on it.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Find the tangent at the right vertex and explain why y=mx+b cannot describe it.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The right vertex is (3,0).

Hint 2

A tangent there is perpendicular to the horizontal radius.

Worked solution

The tangent equation fixes x.
$x=3$
Its slope is undefined, so no finite m works.

x=3.

Checks and common pitfalls: Slope-form parameterizations omit vertical lines.

Reasoning checklist · self / teacher assessment

x=3.
Slope-form parameterizations omit vertical lines.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Find the squared length of the chord cut by y=1.

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

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The endpoints have opposite x-coordinates.

Hint 2

The horizontal difference is twice either positive root.

Worked solution

Find the squared x-coordinate.
$x^2=27/4$
Square the full chord length.
$L^2=4x^2=27$

L²=27.

Checks and common pitfalls: Doubling a length quadruples its square.

Think first. Reveal a hint when the class is ready.

12 / 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.

13 / Transfer#Your turn

A chord of y²=4x has midpoint (2,1). Recover the chord line and both endpoints.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Let the endpoint ordinates be y₁,y₂.

Hint 2

Use their sum and the midpoint x-coordinate to obtain their product.

Worked solution

Midpoint conditions give sum 2 and sum of squares 16.
$y_1+y_2=2,\ y_1^2+y_2^2=16\Rightarrow y_1y_2=-6$
The ordinates solve one quadratic.
$y^2-2y-6=0\Rightarrow y=1\pm\sqrt7$
Use x=y²/4 and match signs.
$(x,y)=(2\pm\sqrt7/2,1\pm\sqrt7),\ 2x-y=3$

Chord line 2x−y=3; endpoints (2±√7/2,1±√7) with matching signs.

Checks and common pitfalls: The two ± signs are linked, not independent choices giving four points.

Reasoning checklist · self / teacher assessment

Chord line 2x−y=3; endpoints (2±√7/2,1±√7) with matching signs.
The two ± signs are linked, not independent choices giving four points.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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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.”