Question

点p(x,y)在圆x2+y2=4上,则y-4/x-4的最大值是

Answer 1

解：点P(x，y)在圆x2 + y2 = 4上，令t = (y – 4)/(x – 4)，则y = t(x – 4) + 4 = tx + 4 – 4t代入圆的方程，可得x2 + (tx + 4 – 4t)2 = 4 => (1 + t2)x2 + 8t(1 – t)x + 16t2 – 32t + 12 = 0 => Δ= 64t2(1 – t)2 – 4(1 + t2)(16t2 – 32t + 12) = 64t2(t2 – 2t + 1) – 16(1 + t2)(4t2 – 8t + 3) = (64t4 – 128t3 + 64t2) – 16(4t4 – 8t3 + 3t2 + 4t2 – 8t + 3) = 64t4 – 128t3 + 64t2 – 64t4 + 128t3 – 112t2 + 128t – 48 = -48t2 + 128t – 48 ≥ 0 => 3t2 – 8t + 3 ≤ 0①，令3t2 – 8t + 3 = 0，可知根的判别式△= 64 – 4*3*3 = 28，对应的一元二次方程的两根为x = (8±√28)/6 = (4±√7)/3，所以不等式①的解集为t∈[(4 - √7)/3，(4 + √7)/3]，所以t = (y – 4)/(x – 4)的最大值是(4 + √7)/3 。

Answer 2

2
望采纳！

Answer 3

答案为4