Kenneth ChengMathematical modeling
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LEARN · EXPLAIN · REVISE

Parabola

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

高二選擇性必修 第一册(A版).pdf · 3.3 · PDF 135 / printed page 130

Revisit first: Hyperbola

TOPIC 01

Parabola

Relate focus, directrix and vertex, and retain conditions in line-parabola intersections.

What you will be able to explain

  • Relate focus, directrix and vertex, and retain conditions in line-parabola intersections.
  • Justify the method and check the conditions in a new situation.

Focus-directrix

y²=4ax has focus (a,0) and directrix x=−a.

Nondegeneracy

a must be nonzero; a parabola has a vertex and no center of symmetry.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: How does changing the sign of a alter y²=4ax?

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

Parabola y²=4a(x−b), a=2≠0. Vertex=(1,0), focus=(3,0); there is no centre of symmetry.

Parabola y²=4a(x−b), a=2≠0. Vertex=(1,0), focus=(3,0); there is no centre of symmetry.

Explain: Calculate two valid cases and explain the change using the defining relation.

Transfer: Translate a parabola and distinguish its vertex from a nonexistent symmetry center.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Worked example

Find the x-coordinate of the focus.

y2=8xy^2=8x
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
4a=4t4a=4t
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    a=2⇒F=(2,0)a=2\Rightarrow F=(2,0)

  3. Read the focal parameter after matching the coefficient 4a.

The requested value is 2.

Checks and common pitfalls: Read the focal parameter after matching the coefficient 4a.

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

02 / Standard#Worked example

Find the length of the focal chord perpendicular to the axis.

y2=12xy^2=12x
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
x=a⇒y2=4a2x=a\Rightarrow y^2=4a^2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    y=±2(3)⇒L=12y=\pm2(3)\Rightarrow L=12

  3. The latus rectum has length 4|a|.

The requested value is 12.

Checks and common pitfalls: The latus rectum has length 4|a|.

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

03 / Transfer#Worked example

Find all k giving exactly one common point.

y2=16x,y=kx+8y^2=16x, y=kx+8
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Handle k=0 before using a discriminant.
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    k=0⇒y=8,x=4k=0 ⇒ y=8, x=4

  3. Apply the stated relation and retain its conditions.

    k≠0:Δ=256(1−2k)k≠0: \Delta=256(1-2k)

  4. Apply the stated relation and retain its conditions.

    Δ=0⇒k=1/2\Delta=0 ⇒ k=1/2

  5. The horizontal case is transverse; the second case is tangent.

The requested relation or conclusion is shown below.

k=0ork=1/2k=0\quad\text{or}\quad k=1/2

Checks and common pitfalls: The horizontal case is transverse; the second case is tangent.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

04 / Foundation#Your turn

Find the x-coordinate of the focus.

y2=20xy^2=20x
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
4a=4t4a=4t
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    a=5⇒F=(5,0)a=5\Rightarrow F=(5,0)

  3. Read the focal parameter after matching the coefficient 4a.

The requested value is 5.

Checks and common pitfalls: Read the focal parameter after matching the coefficient 4a.

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

05 / Foundation#Your turn

Find the x-coordinate of the directrix.

y2=24xy^2=24x
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
directrix:x=−a\text{directrix}:x=-a
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    x=−6x=-6

  3. Focus and directrix are on opposite sides of the vertex.

The requested value is -6.

Checks and common pitfalls: Focus and directrix are on opposite sides of the vertex.

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

06 / Foundation#Your turn

Find the parabola with vertex O and focus (0,−t).

t=7t=7
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
x2=4ay,a=−tx^2=4ay,\quad a=-t
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    x2=4(−7)yx^2=4(-7)y

  3. The focus below the vertex gives a downward opening.

The requested relation or conclusion is shown below.

x2=−28yx^2=-28y

Checks and common pitfalls: The focus below the vertex gives a downward opening.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

07 / Foundation#Your turn

On y²=4tx, P has x-coordinate 2t. Find PF.

t=8t=8
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
PF=dist⁡(P,x=−t)PF=\operatorname{dist}(P,x=-t)
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    PF=∣16+8∣=24PF=|16+8|=24

  3. Use equal distance to focus and directrix without solving y.

The requested value is 24.

Checks and common pitfalls: Use equal distance to focus and directrix without solving y.

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

08 / Standard#Your turn

On y²=4tx, P has x-coordinate 2t. Find PF.

t=9t=9
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
PF=dist⁡(P,x=−t)PF=\operatorname{dist}(P,x=-t)
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    PF=∣18+9∣=27PF=|18+9|=27

  3. Use equal distance to focus and directrix without solving y.

The requested value is 27.

Checks and common pitfalls: Use equal distance to focus and directrix without solving y.

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

09 / Standard#Your turn

Find the length of the focal chord perpendicular to the axis.

y2=40xy^2=40x
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
x=a⇒y2=4a2x=a\Rightarrow y^2=4a^2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    y=±2(10)⇒L=40y=\pm2(10)\Rightarrow L=40

  3. The latus rectum has length 4|a|.

The requested value is 40.

Checks and common pitfalls: The latus rectum has length 4|a|.

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

10 / Standard#Your turn

Find the tangent of slope 1 to y²=4tx.

t=11t=11
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
y=x+b⇒x2+(2b−4t)x+b2=0y=x+b\Rightarrow x^2+(2b-4t)x+b^2=0
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    Δ=(2b−44)2−4b2=176(11−b)\Delta=(2b-44)^2-4b^2=176(11-b)

  3. Apply the stated relation and retain its conditions.

    Δ=0⇒b=11\Delta=0\Rightarrow b=11

  4. The quadratic coefficient remains nonzero, so a repeated root establishes tangency.

The requested relation or conclusion is shown below.

y=x+11y=x+11

Checks and common pitfalls: The quadratic coefficient remains nonzero, so a repeated root establishes tangency.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

11 / Standard#Your turn

Find all k giving exactly one common point.

y2=48x,y=kx+24y^2=48x, y=kx+24
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Handle k=0 before using a discriminant.
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    k=0⇒y=24,x=12k=0 ⇒ y=24, x=12

  3. Apply the stated relation and retain its conditions.

    k≠0:Δ=2304(1−2k)k≠0: \Delta=2304(1-2k)

  4. Apply the stated relation and retain its conditions.

    Δ=0⇒k=1/2\Delta=0 ⇒ k=1/2

  5. The horizontal case is transverse; the second case is tangent.

The requested relation or conclusion is shown below.

k=0ork=1/2k=0\quad\text{or}\quad k=1/2

Checks and common pitfalls: The horizontal case is transverse; the second case is tangent.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

12 / Transfer#Your turn

Find the tangent of slope 1 to y²=4tx.

t=13t=13
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
y=x+b⇒x2+(2b−4t)x+b2=0y=x+b\Rightarrow x^2+(2b-4t)x+b^2=0
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    Δ=(2b−52)2−4b2=208(13−b)\Delta=(2b-52)^2-4b^2=208(13-b)

  3. Apply the stated relation and retain its conditions.

    Δ=0⇒b=13\Delta=0\Rightarrow b=13

  4. The quadratic coefficient remains nonzero, so a repeated root establishes tangency.

The requested relation or conclusion is shown below.

y=x+13y=x+13

Checks and common pitfalls: The quadratic coefficient remains nonzero, so a repeated root establishes tangency.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

13 / Transfer#Your turn

Find all k giving exactly one common point.

y2=56x,y=kx+28y^2=56x, y=kx+28
  • a must be nonzero; a parabola has a vertex and no center of symmetry.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Handle k=0 before using a discriminant.
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    k=0⇒y=28,x=14k=0 ⇒ y=28, x=14

  3. Apply the stated relation and retain its conditions.

    k≠0:Δ=3136(1−2k)k≠0: \Delta=3136(1-2k)

  4. Apply the stated relation and retain its conditions.

    Δ=0⇒k=1/2\Delta=0 ⇒ k=1/2

  5. The horizontal case is transverse; the second case is tangent.

The requested relation or conclusion is shown below.

k=0ork=1/2k=0\quad\text{or}\quad k=1/2

Checks and common pitfalls: The horizontal case is transverse; the second case is tangent.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

Focus on one question

End-of-lesson check and correction

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    Teacher preparation and assessment

    Question sequence

    • Relate focus, directrix and vertex, and retain conditions in line-parabola intersections.
    • Which condition is essential in parabola?
    • How does changing the sign of a alter y²=4ax?

    Board plan

    • Focus-directrix: y²=4ax has focus (a,0) and directrix x=−a.
    • Nondegeneracy: a must be nonzero; a parabola has a vertex and no center of symmetry.

    Anticipated thinking

    • For y²=2px the focal coordinate is p/2, not p.

    Assessment checklist

    • 1 mark: choose the correct representation and conditions.
    • 1 mark: establish the intermediate relation.
    • 1 mark: complete a connected calculation or proof.
    • 1 mark: interpret and check the conclusion.

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    Curriculum and source notes ↗