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Find slopes for which OA⊥OB, where A,B are the intersections and O is the origin.

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

TOPIC 01

2024 JM02

The denominator restriction m≠±2 remains in force.

Try it. Leave your reasoning visible.

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

01 / Standard#Your turn

Find slopes for which OA⊥OB, where A,B are the intersections and O is the origin.

H:x2−y24=1,L:y=m(x−5)H:x^2-\frac{y^2}{4}=1,\quad L:y=m(x-\sqrt5)

Official paper · jm02-2024 · 3(c) · PDF 5

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the root sum and product from part (a).
Hint 2
Set the position-vector dot product equal to zero.
Worked solution
  1. Read Vieta’s formulae.

    s=x1+x2=25m2m2−4,p=x1x2=5m2+4m2−4s=x_1+x_2=\frac{2\sqrt5m^2}{m^2-4},\quad p=x_1x_2=\frac{5m^2+4}{m^2-4}
  2. Expand the dot product and substitute.

    0=p+m2(p−5s+5)=4−11m2m2−40=p+m^2(p-\sqrt5s+5)=\frac{4-11m^2}{m^2-4}
  3. Solve and check the excluded values.

    m=±211=±21111≠±2m=\pm\frac2{\sqrt{11}}=\pm\frac{2\sqrt{11}}{11}\ne\pm2

m=±2√11/11.

Checks and common pitfalls: The denominator restriction m≠±2 remains in force.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Read Vieta’s formulae.
    s=x1+x2=25m2m2−4,p=x1x2=5m2+4m2−4s=x_1+x_2=\frac{2\sqrt5m^2}{m^2-4},\quad p=x_1x_2=\frac{5m^2+4}{m^2-4}
  • Expand the dot product and substitute.
    0=p+m2(p−5s+5)=4−11m2m2−40=p+m^2(p-\sqrt5s+5)=\frac{4-11m^2}{m^2-4}
  • Solve and check the excluded values.
    m=±211=±21111≠±2m=\pm\frac2{\sqrt{11}}=\pm\frac{2\sqrt{11}}{11}\ne\pm2

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Curriculum and source notes ↗