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If y=mx+c is tangent to x²=8y, prove c=−2m².

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

TOPIC 01

2021 JM02

Here the elimination remains quadratic, so the double-root criterion is valid even when m=0.

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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 y=mx+c is tangent to x²=8y, prove c=−2m².

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

Official paper · jm02-2021 · 3(a) · PDF 5

Official original and suggested answers ↗ · Suggested answer PDF page 9

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Substitute the line into the parabola.
Hint 2
The intersection quadratic has a repeated root at tangency.
Worked solution
  1. The leading quadratic coefficient remains one for every m.

    x2−8mx−8c=0x^2-8mx-8c=0
  2. Set its discriminant to zero and rearrange.

    Δ=64m2+32c=0  ⟹  c=−2m2\Delta=64m^2+32c=0\implies c=-2m^2

c=−2m².

Checks and common pitfalls: Here the elimination remains quadratic, so the double-root criterion is valid even when m=0.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The leading quadratic coefficient remains one for every m.
    x2−8mx−8c=0x^2-8mx-8c=0
  • Set its discriminant to zero and rearrange.
    Δ=64m2+32c=0  ⟹  c=−2m2\Delta=64m^2+32c=0\implies c=-2m^2

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Curriculum and source notes ↗