The line y=x−6 meets this hyperbola at A,B. With M=(2,0), prove AM is perpendicular to BM.

01 / Standard#Your turn

Official paper · jm01-2026 · II.4(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Eliminate y to obtain a quadratic for the intersections.

Hint 2

Use the root sum and product in the product of slopes.

Worked solution

Substitution gives two real roots and their symmetric sums.
$x^2-2(x-6)^2=4\implies x^2-24x+76=0,\quad x_1+x_2=24,\ x_1x_2=76$
Neither root equals 2, so both slopes exist.
$m_1m_2=\frac{(x_1-6)(x_2-6)}{(x_1-2)(x_2-2)}=\frac{76-6(24)+36}{76-2(24)+4}=-1$
A slope product of −1 proves perpendicularity.

AM⊥BM.

Checks and common pitfalls: The two slope denominators use x₁−2 and x₂−2 respectively.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Substitution gives two real roots and their symmetric sums.
$x^2-2(x-6)^2=4\implies x^2-24x+76=0,\quad x_1+x_2=24,\ x_1x_2=76$
Neither root equals 2, so both slopes exist.
$m_1m_2=\frac{(x_1-6)(x_2-6)}{(x_1-2)(x_2-2)}=\frac{76-6(24)+36}{76-2(24)+4}=-1$
A slope product of −1 proves perpendicularity.

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

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