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A point on the downward parabola has focal distance 15 and distance 7 from the x-axis. Find p.

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

TOPIC 01

2025 JM01

The coefficient p here is four times the focal distance parameter.

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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 point on the downward parabola has focal distance 15 and distance 7 from the x-axis. Find p.

x2=−py,p>0x^2=-py,\quad p>0

Official paper · jm01-2025 · I.11 · PDF 3

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

Skills and prerequisite lessons
  1. Option A1414
  2. Option B1515
  3. Option C1616
  4. Option D2828
  5. Option E3232

Working and explanation

BUILD THE REASONING

Hint 1
The point’s ordinate must be −7.
Hint 2
Its distance to the directrix equals its focal distance.
Worked solution
  1. Identify the directrix using the standard coefficient 4a.

    ydirectrix=p4,yA=−7y_{\rm directrix}=\frac p4,\quad y_A=-7
  2. Apply the focus-directrix definition.

    7+p4=15  ⟹  p=327+\frac p4=15\implies p=32

E: 32.

Checks and common pitfalls: The coefficient p here is four times the focal distance parameter.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Identify the directrix using the standard coefficient 4a.
    ydirectrix=p4,yA=−7y_{\rm directrix}=\frac p4,\quad y_A=-7
  • Apply the focus-directrix definition.
    7+p4=15  ⟹  p=327+\frac p4=15\implies p=32

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Curriculum and source notes ↗