Question 4

**Locus S:** From the two lines, 4x − 3y = 12α and 4x + 3y = 12/α. Multiplying: (4x − 3y)(4x + 3y) = 144 ⟹ 16x² − 9y² = 144 ⟹ x²/9 − y²/16 = 1 (hyperbola, a² = 9, b² = 16). **Tangent T:** Parallel to 4x − (3/√2)y = 0, so slope m = 4√2/3. Tangent to hyperbola with slope m: y = mx ± √(a²m² − b²) = mx ± √(9·32/9 − 16) = mx ± √16 = mx ± 4. Since T passes through (0, q) with q > 0, we take q = 4, so T: y = (4√2/3)x + 4. x-intercept: 0 = (4√2/3)p + 4 ⟹ p = −3/√2. Therefore pq = (−3/√2)(4) = −12/√2 = −6√2.

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