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The angle between the two tangents is π/4 and m₁=2. Find all possible A.

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TOPIC 01

2021 JM02

Using a signed angle formula without the absolute value would lose one of the two unoriented-line configurations.

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01 / Standard#Your turn

The angle between the two tangents is π/4 and m₁=2. Find all possible A.

P:x2=8yP:x^2=8y

Official paper · jm02-2021 · 3(b)(iii) · PDF 5

Official original and suggested answers ↗ · Suggested answer PDF page 10

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the absolute tangent of the angle between the lines.
Hint 2
Retain both solutions after squaring and check the denominator.
Worked solution
  1. Perpendicular slopes are excluded by the specified π/4 angle.

    1=∣2−m21+2m2∣,m2≠−1/21=\left|\frac{2-m_2}{1+2m_2}\right|,\quad m_2\ne-1/2
  2. Squaring and factoring gives two valid slopes.

    (2−m2)2=(1+2m2)2  ⟺  3m22+8m2−3=0  ⟹  m2=−3,1/3(2-m_2)^2=(1+2m_2)^2\iff3m_2^2+8m_2-3=0\implies m_2=-3,1/3
  3. Use the sum and product formulas from part (i).

    A=(−2,−12)orA=(14/3,4/3)A=(-2,-12)\quad\text{or}\quad A=(14/3,4/3)

A=(−2,−12) or (14/3,4/3).

Checks and common pitfalls: Using a signed angle formula without the absolute value would lose one of the two unoriented-line configurations.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Perpendicular slopes are excluded by the specified π/4 angle.
    1=∣2−m21+2m2∣,m2≠−1/21=\left|\frac{2-m_2}{1+2m_2}\right|,\quad m_2\ne-1/2
  • Squaring and factoring gives two valid slopes.
    (2−m2)2=(1+2m2)2  ⟺  3m22+8m2−3=0  ⟹  m2=−3,1/3(2-m_2)^2=(1+2m_2)^2\iff3m_2^2+8m_2-3=0\implies m_2=-3,1/3
  • Use the sum and product formulas from part (i).
    A=(−2,−12)orA=(14/3,4/3)A=(-2,-12)\quad\text{or}\quad A=(14/3,4/3)

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Curriculum and source notes ↗