Zadanie 2c, Paper 3, listopad 2023, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — listopad 2023, Paper 3 (zadania wzorowane), pytanie 2(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
This investigation concerns intersecting curve families. In 2a, take parabolas y=kx², where k≠0; work locally where x,y≠0. In 2b–2d use Ea : x2/(a2 + 9) + y2/a2 = 1 for a>0 and Hb : x2/(9 − b2) − y2/b2 = 1 for 0<b<3. In 2e–2f a tangent of finite slope f is rotated anticlockwise through an angle α; where its new slope is finite, .
Polecenie
Solve E_a and H_b simultaneously to show that their four intersections have coordinates , where the two signs are independent.
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Odpowiedź z klucza
Put U=x²,V=y² and clear denominators: a²U+(a²+9)V=a²(a²+9), b²U−(9−b²)V=b²(9−b²). Multiply the first by b² and the second by a², then subtract. This gives 9(a²+b²)V=a²b²(a²+b²). Since a²+b²>0, V=a²b²/9. Substitution into either equation gives U=(a²+9)(9−b²)/9. Both U and V are positive under the parameter restrictions. Taking independent positive/negative square roots gives all four stated points.
Schemat punktowania
A1:correct equations in U,V or an equivalent elimination setup. M1:eliminate one squared coordinate. R1:justify division by a²+b²>0 or an equivalent nonzero factor. A1:V=a²b²/9. M1:substitute to find U. A1:correct U and the four independent-sign coordinate pairs, establishing the target. The marks after the nonzero-divisor R1 are independent of that R1. Accept elimination in the reverse order (first U, then V) with equivalent six-mark allocation.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Exact claims and proofs require analytical working. Counts of intercepts mean distinct real points, unless repeated roots are explicitly requested.