Find the locus equation, retaining the condition that both slopes exist.

01 / Standard#Your turn

A=(-2\sqrt2,0),\ B=(2\sqrt2,0),\quad m_{AM}m_{BM}=-\frac12

Official paper · jm01-2025 · II.5(a) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Write each slope from its two coordinates.

Hint 2

Record the excluded vertical-line abscissas before multiplying denominators.

Worked solution

Both slopes exist only away from the two endpoint abscissas.
$\frac y{x+2\sqrt2}\frac y{x-2\sqrt2}=-\frac12,\quad x\ne\pm2\sqrt2$
Clear the nonzero denominator and rearrange.
$2y^2=8-x^2\implies\frac{x^2}8+\frac{y^2}4=1,\quad M\ne(\pm2\sqrt2,0)$

Ellipse x²/8+y²/4=1 with the two vertices (±2√2,0) excluded under the literal slope condition.

Checks and common pitfalls: Clearing denominators can add points at which the original slopes were undefined.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Both slopes exist only away from the two endpoint abscissas.
$\frac y{x+2\sqrt2}\frac y{x-2\sqrt2}=-\frac12,\quad x\ne\pm2\sqrt2$
Clear the nonzero denominator and rearrange.
$2y^2=8-x^2\implies\frac{x^2}8+\frac{y^2}4=1,\quad M\ne(\pm2\sqrt2,0)$

Think first. Reveal a hint when the class is ready.

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