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If the two tangents in part (b) are perpendicular, find the complete locus of A.

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

TOPIC 01

2021 JM02

A necessary equation for a locus also needs a sufficiency argument.

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

01 / Standard#Your turn

If the two tangents in part (b) are perpendicular, find the complete locus of A.

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

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

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Both slopes are finite, so m₁m₂=−1.
Hint 2
Show every real h occurs, not merely that k=−2.
Worked solution
  1. The ordinate is fixed by the slope product.

    k=2m1m2=−2k=2m_1m_2=-2
  2. For any real h, solve the tangent-slope equation.

    2m2−hm−2=0,Δ=h2+16>02m^2-hm-2=0,\quad\Delta=h^2+16>0
  3. Its two distinct real roots have product −1, so their tangents meet at (h,−2) and are perpendicular.

    m1+m2=h/2,m1m2=−1m_1+m_2=h/2,\quad m_1m_2=-1

The full straight line y=−2.

Checks and common pitfalls: A necessary equation for a locus also needs a sufficiency argument.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The ordinate is fixed by the slope product.
    k=2m1m2=−2k=2m_1m_2=-2
  • For any real h, solve the tangent-slope equation.
    2m2−hm−2=0,Δ=h2+16>02m^2-hm-2=0,\quad\Delta=h^2+16>0
  • Its two distinct real roots have product −1, so their tangents meet at (h,−2) and are perpendicular.
    m1+m2=h/2,m1m2=−1m_1+m_2=h/2,\quad m_1m_2=-1

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Curriculum and source notes ↗