Kenneth ChengMathematical modeling
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Conic sections: mixed review

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

TOPIC 01

Conic sections: mixed review

A mixed assessment: identify the method, justify it and revise your reasoning.

What you will be able to explain

  • Connect the chapter skills without relying on the order of the exercises.

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

Find the major-axis length.

x2/484+y2/400=1x^2/484+y^2/400=1
  • For a horizontal major axis a>b>0; the larger denominator specifies the major axis.
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.
major axis=2a\text{major axis}=2a
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    2a=2(22)=442a=2(22)=44

  3. An axis is twice the corresponding semiaxis.

The requested value is 44.

Checks and common pitfalls: An axis is twice the corresponding semiaxis.

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

02 / Foundation#Your turn

Find the length of the transverse axis.

x2/441−y2/4=1x^2/441-y^2/4=1
  • a,b>0 and e>1; a distance difference is absolute, and an asymptote is not part of the curve.
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.
transverse axis=2a\text{transverse axis}=2a
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    2a=422a=42

  3. Only the positive squared term specifies the transverse axis.

The requested value is 42.

Checks and common pitfalls: Only the positive squared term specifies the transverse axis.

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

03 / Foundation#Your turn

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

t=22t=22
  • 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(−22)yx^2=4(-22)y

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

The requested relation or conclusion is shown below.

x2=−88yx^2=-88y

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.

04 / Foundation#Your turn

P is on the ellipse and PF1=t. Find PF2.

a=25,PF1=23a=25,\quad PF_1=23
  • For a horizontal major axis a>b>0; the larger denominator specifies the major axis.
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.
PF1+PF2=2aPF_1+PF_2=2a
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    PF2=50−23=27PF_2=50-23=27

  3. The distance-sum definition avoids solving coordinates.

The requested value is 27.

Checks and common pitfalls: The distance-sum definition avoids solving coordinates.

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

05 / Foundation#Your turn

Find c².

x2/576−y2/4=1x^2/576-y^2/4=1
  • a,b>0 and e>1; a distance difference is absolute, and an asymptote is not part of the curve.
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.
c2=a2+b2c^2=a^2+b^2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    c2=576+4=580c^2=576+4=580

  3. The focal square is a sum for a hyperbola.

The requested value is 580.

Checks and common pitfalls: The focal square is a sum for a hyperbola.

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

06 / Foundation#Your turn

Find the x-coordinate of the directrix.

y2=100xy^2=100x
  • 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=−25x=-25

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

The requested value is -25.

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

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

07 / Standard#Your turn

P is on the ellipse and PF1=t. Find PF2.

a=28,PF1=26a=28,\quad PF_1=26
  • For a horizontal major axis a>b>0; the larger denominator specifies the major axis.
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.
PF1+PF2=2aPF_1+PF_2=2a
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    PF2=56−26=30PF_2=56-26=30

  3. The distance-sum definition avoids solving coordinates.

The requested value is 30.

Checks and common pitfalls: The distance-sum definition avoids solving coordinates.

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

08 / Standard#Your turn

Find the horizontal hyperbola with vertices (±t,0) and foci (±(t+2),0).

t=27t=27
  • a,b>0 and e>1; a distance difference is absolute, and an asymptote is not part of the curve.
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.
b2=c2−a2b²=c²−a²
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    b2=841−729=112b^2=841-729=112

  3. For a hyperbola c exceeds a.

The requested relation or conclusion is shown below.

x2/729−y2/112=1x^2/729-y^2/112=1

Checks and common pitfalls: For a hyperbola c exceeds a.

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.

09 / Standard#Your turn

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

t=28t=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
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−112)2−4b2=448(28−b)\Delta=(2b-112)^2-4b^2=448(28-b)

  3. Apply the stated relation and retain its conditions.

    Δ=0⇒b=28\Delta=0\Rightarrow b=28

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

The requested relation or conclusion is shown below.

y=x+28y=x+28

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.

10 / Standard#Your turn

P is on the ellipse and PF1=t. Find PF2.

a=31,PF1=29a=31,\quad PF_1=29
  • For a horizontal major axis a>b>0; the larger denominator specifies the major axis.
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.
PF1+PF2=2aPF_1+PF_2=2a
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    PF2=62−29=33PF_2=62-29=33

  3. The distance-sum definition avoids solving coordinates.

The requested value is 33.

Checks and common pitfalls: The distance-sum definition avoids solving coordinates.

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

11 / Standard#Your turn

Find the horizontal hyperbola with vertices (±t,0) and foci (±(t+2),0).

t=30t=30
  • a,b>0 and e>1; a distance difference is absolute, and an asymptote is not part of the curve.
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.
b2=c2−a2b²=c²−a²
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    b2=1024−900=124b^2=1024-900=124

  3. For a hyperbola c exceeds a.

The requested relation or conclusion is shown below.

x2/900−y2/124=1x^2/900-y^2/124=1

Checks and common pitfalls: For a hyperbola c exceeds a.

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 / Standard#Your turn

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

t=31t=31
  • 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−124)2−4b2=496(31−b)\Delta=(2b-124)^2-4b^2=496(31-b)

  3. Apply the stated relation and retain its conditions.

    Δ=0⇒b=31\Delta=0\Rightarrow b=31

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

The requested relation or conclusion is shown below.

y=x+31y=x+31

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 the length of the chord cut by x=0.

x2/1156+y2/1024=1x^2/1156+y^2/1024=1
  • For a horizontal major axis a>b>0; the larger denominator specifies the major axis.
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.
y2/b2=1y^2/b^2=1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    y=±32⇒L=64y=\pm32\Rightarrow L=64

  3. The chord through the center along the y-axis is the minor axis.

The requested value is 64.

Checks and common pitfalls: The chord through the center along the y-axis is the minor axis.

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

14 / Transfer#Your turn

Show there is one common point and explain why this line is not tangent.

y=x+33,x2−y2=1089y=x+33, x^2-y^2=1089
  • a,b>0 and e>1; a distance difference is absolute, and an asymptote is not part of the curve.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Substitution cancels the quadratic terms.
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    −66x−1089=1089⇒x=−33,y=0-66x-1089=1089 ⇒ x=-33, y=0

  3. Apply the stated relation and retain its conditions.

    tangent at P:x=−33\text{tangent at }P:x=-33

  4. The line is parallel to an asymptote; one intersection can result from degree reduction.

The requested relation or conclusion is shown below.

P=(−33,0)P=(-33,0)

Checks and common pitfalls: The line is parallel to an asymptote; one intersection can result from degree reduction.

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.

15 / Transfer#Your turn

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

t=34t=34
  • 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−136)2−4b2=544(34−b)\Delta=(2b-136)^2-4b^2=544(34-b)

  3. Apply the stated relation and retain its conditions.

    Δ=0⇒b=34\Delta=0\Rightarrow b=34

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

The requested relation or conclusion is shown below.

y=x+34y=x+34

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.

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    Teacher preparation and assessment

    Question sequence

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    Board plan

    • Compare valid methods and annotate their conditions.

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    • A correct final value may still hide a missing assumption.

    Assessment checklist

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