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A horizontal hyperbola has eccentricity √10/2 and contains (2√2,3). Find its equation.

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

TOPIC 01

2024 JM01

For a hyperbola, c²=a²+b².

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

01 / Standard#Your turn

A horizontal hyperbola has eccentricity √10/2 and contains (2√2,3). Find its equation.

x2a2−y2b2=1,a,b>0\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\quad a,b>0

Official paper · jm01-2024 · II.5(a) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use e²=1+b²/a².
Hint 2
Substitute the given point.
Worked solution
  1. Relate the two squared parameters.

    b2a2=52−1=32\frac{b^2}{a^2}=\frac52-1=\frac32
  2. Solve using the point condition.

    8a2−9(3/2)a2=1  ⟹  a2=2, b2=3\frac8{a^2}-\frac9{(3/2)a^2}=1\implies a^2=2,\ b^2=3

x²/2−y²/3=1.

Checks and common pitfalls: For a hyperbola, c²=a²+b².

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Relate the two squared parameters.
    b2a2=52−1=32\frac{b^2}{a^2}=\frac52-1=\frac32
  • Solve using the point condition.
    8a2−9(3/2)a2=1  ⟹  a2=2, b2=3\frac8{a^2}-\frac9{(3/2)a^2}=1\implies a^2=2,\ b^2=3

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Curriculum and source notes ↗