[Toán 12] Phương pháp toạ độ trong mặt phẳng

N

nguyenbahiep1

[laTEX]y = k(x-1) + \frac{1}{2} \\ \\ \frac{x^2}{4} + k^2(x-1)^2 + k(x-1) + \frac{1}{4} = 1 \\ \\ x^2 + 4k^2(x-1)^2 + 4k(x-1) - 3 = 0 \\ \\ (4k^2+1).x^2 - 2(4k^2 -2k)x + 4k^2 -4k-3 = 0 \\ \\ \Delta'= (4k^2-2k)^2 - 4(4k^2+1)(4k^2-4k-3) > 0 \\ \\ (x_A +x_B):2 = \frac{4k^2 -2k}{4k^2+1} = 1 \Rightarrow k = - \frac{1}{2} \Rightarrow \Delta' =4 > 0 (T/M) \\ \\ (y_A + y_B ): 2 = \frac{1}{2} \\ \\ (d): y = -\frac{1}{2}.x + 1 [/laTEX]
 
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