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Intermediate Algebra / LEVEL 3 · DIFFICULTY 3/5

Conics, Series and Bounds

Recognize common algebraic structures.

3 stages · 24 practice problems · two 6-question assessment forms

Choose an island to read its lesson.

  1. MINI QUEST IA 3.1Ellipse ParametersRead the lesson
  2. MINI QUEST IA 3.2Finite Geometric SumsRead the lesson
  3. MINI QUEST IA 3.3Bounding Positive ExpressionsRead the lesson
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STAGE IA 3.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Ellipse Parameters

Useful preparation: Factors and Unknown Coefficients

Goal: Understand and apply ellipse parameters.

Before you begin: Factors and Unknown Coefficients

Understand the idea

A standard ellipse centered at the origin has two semiaxis lengths. Its foci lie on the longer axis, and the squared focal distance is the difference of the squared semiaxis lengths.

x²/a²+y²/b²=1, c²=a²−b² when a≥b

Choose and carry out a method

Identify the larger denominator as a² and the smaller as b². Compute c²=a²−b² when the question requests the square of the focal distance.

Check the reasoning

The focus lies inside the ellipse, so c<a. Equal semiaxes produce a circle and focal distance zero.

WORKED EXAMPLE 1

An ellipse has equation x²/36+y²/9=1. Find c², where its foci are (±c,0).

  1. For an ellipse, the focal distance satisfies c²=a²-b².
  2. The larger denominator is a²=36; the smaller is b²=9.
  3. Thus c²=27. The foci lie inside the major-axis vertices.

27

WORKED EXAMPLE 2

An ellipse has equation x²/49+y²/16=1. Find c², where its foci are (±c,0).

  1. For an ellipse, the focal distance satisfies c²=a²-b².
  2. The larger denominator is a²=49; the smaller is b²=16.
  3. Thus c²=33. The foci lie inside the major-axis vertices.

33

Common pitfalls

Possible mix-up: Add the semiaxis squares to find c².

For an ellipse, use their difference.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

What happens to the foci as the two semiaxes approach equality?

Preview the eight practice prompts
  1. An ellipse has equation x²/81+y²/36=1. Find c², where its foci are (±c,0).
  2. An ellipse has equation x²/100+y²/49=1. Find c², where its foci are (±c,0).
  3. An ellipse has equation x²/121+y²/64=1. Find c², where its foci are (±c,0).
  4. An ellipse has equation x²/144+y²/81=1. Find c², where its foci are (±c,0).
  5. An ellipse has equation x²/169+y²/100=1. Find c², where its foci are (±c,0).
  6. An ellipse has equation x²/196+y²/121=1. Find c², where its foci are (±c,0).
  7. An elliptical mirror has semimajor axis 15 cm and semiminor axis 12 cm. What is the square of the distance from its center to either focus? New context
  8. An elliptical mirror has semimajor axis 16 cm and semiminor axis 13 cm. What is the square of the distance from its center to either focus? New context
Open stage IA 3.1 in the student workspace →

STAGE IA 3.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Finite Geometric Sums

Useful preparation: Ellipse Parameters

Goal: Understand and apply finite geometric sums.

Before you begin: Ellipse Parameters

Understand the idea

A geometric sequence multiplies each term by a fixed ratio. Multiplying its sum by that ratio shifts the terms so subtraction cancels every interior term.

Sₙ=a(rⁿ−1)/(r−1), r≠1

Choose and carry out a method

Write S and rS with aligned powers, subtract, then divide by r−1 when r≠1. Count the number of terms carefully.

Check the reasoning

The first term uses exponent zero. When r=1, use n times the common term instead of a formula with a zero denominator.

WORKED EXAMPLE 1

Find the sum of the first 5 terms of 3, 6, 12, … .

  1. Subtract the original sum from twice the sum to cancel all middle terms.
  2. S=3(1+2+…+2^4)=3(2^5-1).
  3. The total is 93; the final term alone is only part of the sum.

93

WORKED EXAMPLE 2

Find the sum of the first 6 terms of 4, 8, 16, … .

  1. Subtract the original sum from twice the sum to cancel all middle terms.
  2. S=4(1+2+…+2^5)=4(2^6-1).
  3. The total is 252; the final term alone is only part of the sum.

252

Common pitfalls

Possible mix-up: The last exponent equals the number of terms.

Starting with exponent zero makes the last exponent n−1.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Derive the sum formula by cancellation rather than memorization.

Preview the eight practice prompts
  1. Find the sum of the first 4 terms of 6, 12, 24, … .
  2. Find the sum of the first 5 terms of 7, 14, 28, … .
  3. Find the sum of the first 6 terms of 8, 16, 32, … .
  4. Find the sum of the first 7 terms of 9, 18, 36, … .
  5. Find the sum of the first 4 terms of 10, 20, 40, … .
  6. Find the sum of the first 5 terms of 11, 22, 44, … .
  7. A message chain sends 12 messages in its first round and doubles the number each round. How many messages are sent during the first 6 rounds altogether? New context
  8. A message chain sends 13 messages in its first round and doubles the number each round. How many messages are sent during the first 7 rounds altogether? New context
Open stage IA 3.2 in the student workspace →

STAGE IA 3.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Bounding Positive Expressions

Useful preparation: Finite Geometric Sums

Goal: Understand and apply bounding positive expressions.

Before you begin: Finite Geometric Sums

Understand the idea

For nonnegative x and y, (√x−√y)²≥0 rearranges to x+y≥2√xy. A fixed product therefore gives a lower bound for the sum.

x+y≥2√xy, equality when x=y

Choose and carry out a method

Apply the inequality and identify its equality condition x=y. Show that the allowed values can actually meet that condition before claiming a minimum.

Check the reasoning

A bound alone need not be attainable under extra restrictions. Check positivity and every constraint in the problem.

WORKED EXAMPLE 1

Positive real numbers x and y satisfy xy=16. Find the minimum possible x+y.

  1. For positive numbers, x+y ≥ 2√(xy), with equality exactly when x=y.
  2. x+y≥2√16=8. Choose x=y=4.
  3. The minimum is attained and equals 8. This is the semiperimeter, not the full perimeter.

8

WORKED EXAMPLE 2

Positive real numbers x and y satisfy xy=25. Find the minimum possible x+y.

  1. For positive numbers, x+y ≥ 2√(xy), with equality exactly when x=y.
  2. x+y≥2√25=10. Choose x=y=5.
  3. The minimum is attained and equals 10. This is the semiperimeter, not the full perimeter.

10

Common pitfalls

Possible mix-up: An inequality bound automatically occurs.

Verify an allowed equality case to establish the minimum.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Why does making two positive factors more equal reduce their sum at fixed product?

Preview the eight practice prompts
  1. Positive real numbers x and y satisfy xy=49. Find the minimum possible x+y.
  2. Positive real numbers x and y satisfy xy=64. Find the minimum possible x+y.
  3. Positive real numbers x and y satisfy xy=81. Find the minimum possible x+y.
  4. Positive real numbers x and y satisfy xy=100. Find the minimum possible x+y.
  5. Positive real numbers x and y satisfy xy=121. Find the minimum possible x+y.
  6. Positive real numbers x and y satisfy xy=144. Find the minimum possible x+y.
  7. A rectangle has area 169 square units. What is the smallest possible sum of its length and width? New context
  8. A rectangle has area 196 square units. What is the smallest possible sum of its length and width? New context
Open stage IA 3.3 in the student workspace →
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