若实数x,y满足x²+y²=1,求y-1÷x-1的最小值,求(x-2)²+(y-1)²的范围?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 01:25:39
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若实数x,y满足x²+y²=1,求y-1÷x-1的最小值,求(x-2)²+(y-1)²的范围?
若实数x,y满足x²+y²=1,求y-1÷x-1的最小值,求(x-2)²+(y-1)²的范围?
若实数x,y满足x²+y²=1,求y-1÷x-1的最小值,求(x-2)²+(y-1)²的范围?
设(y-1)/(x-1)=k,则有y=k(x-1)+1=kx-k+1
代入到圆方程中有x^2+(kx-k+1)^2=1
(1+k^2)x^2-2k(k-1)x+(k-1)^2-1=0
判别式=4k^2(k-1)^2-4(1+k^2)*[(k-1)^2-1]>=0
[(k-1)^2*(4k^2-4-4k^2)+4(1+k^2)>=0
-4(k-1)^2+4+4k^2>=0
-4k^2+8k-4+4+4k^2>=0
k>=0
即有(y-1)/(x-1)的最小值是:0
(X-2)^2+(Y-1)^2表示圆上一点到(2,1)的距离是平方.
最小距离的平方=[根号(2^2+1^2)-1]^2=5+1-2根号5=6-2根号5
最大距离的平方=[根号5+1]^2=6+2根号5
所以,范围是[6-2根号5,6+2根号5]