Given that hyperbola y = 3x and straight line y = KX + 2 intersect point a (x1, Y1) and point B (X2, Y2), and X12 + X22 = 10, find the value of K
From y = KX + 2Y = 3x, it is obtained that 3x = KX + 2, kx2 + 2x-3 = 0. Therefore, X12 + X22 = (x1 + x2) 2-2x1 · x2 = 4k2 + 6K = 10. 5k2-3k-2 = 0, K1 = 1 or K2 = - 25. (4 points) and △ = 4 + 12K > 0, that is, K > - 13, so the value of K is 1. (6 points)
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