Zadanie 1f, Paper 3, listopad 2025, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2025, TZ1, Paper 3 (zadania wzorowane), pytanie 1(f). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
In parts 1a–1d consider H : y = −9/x, x≠0. Its normal at x=t, t≠0, is written y=mx+c. We study normal lines that pass through a fixed point (0,c), rather than normal lines with a prescribed slope. Count geometrically distinct lines: one line may be normal at two different points. In parts 1e–1f use F : y = −k2/x, where k>0. Define A=(kt,−k/t), with t≠0,±1. Its normal meets F again at B. The line OA, where O=(0,0), meets F again at C.
Polecenie
Prove that angle BCA is a right angle.
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The second intersection ofOA with F is C=(−kt,k/t), by central symmetry or direct substitution in y=−x/t². Then CA=(2kt,−2k/t) and CB=(kt−k/t³,kt³−k/t). Their dot product is 2k²t²−2k²/t²−2k²t²+2k²/t²=0. Becausek>0,t≠0,±1, both vectors are nonzero, proving the right angle. Equivalently slopeCA=−1/t² and slopeCB=t², with product −1. A proof by constructing the perpendicular at C and showingB lies on it, or by the converse of Pythagoras, is also valid.
Schemat punktowania
A1:C=(−kt,k/t). M1:form a relevant side vector or gradient. A1:both correct vectors/gradients. M1:use a dot product, product of slopes or equivalent perpendicular test. A1:correct simplification to zero dot product or −1 gradient product. R1:deduce the right angle, with the excluded degenerate values respected. Equivalent perpendicular-line or Pythagorean proofs receive these six marks.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Show complete reasoning and reproducible numerical work. Give exact results unless a stated approximation is requested.