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The hyperbola has eccentricity 3. Minimise (b²+2)/a for a,b>0.

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

TOPIC 01

2023 JM01

The numerator is b²+2 as a whole.

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

01 / Standard#Your turn

The hyperbola has eccentricity 3. Minimise (b²+2)/a for a,b>0.

x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

Official paper · jm01-2023 · I.12 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A22
  2. Option B222\sqrt2
  3. Option C232\sqrt3
  4. Option D44
  5. Option E88

Working and explanation

BUILD THE REASONING

Hint 1
Convert eccentricity into a relation between a and b.
Hint 2
Apply AM–GM to two positive terms.
Worked solution
  1. Use c²=a²+b² and c/a=3.

    b2=8a2,b2+2a=8a+2ab^2=8a^2,\quad\frac{b^2+2}{a}=8a+\frac2a
  2. The product is constant and equality is attainable.

    8a+2/a≥216=8,8a=2/a  ⟺  a=1/28a+2/a\ge2\sqrt{16}=8,\quad8a=2/a\iff a=1/2

E: 8.

Checks and common pitfalls: The numerator is b²+2 as a whole.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Use c²=a²+b² and c/a=3.
    b2=8a2,b2+2a=8a+2ab^2=8a^2,\quad\frac{b^2+2}{a}=8a+\frac2a
  • The product is constant and equality is attainable.
    8a+2/a≥216=8,8a=2/a  ⟺  a=1/28a+2/a\ge2\sqrt{16}=8,\quad8a=2/a\iff a=1/2

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Curriculum and source notes ↗