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构造直线和圆有交点,利用点线距离公式可以简洁地解答不少问题. 例1若实数x,y适合方程x2+y2-2x-4y +1=0.那么代数式y/x+2的取值范围是____. 解:令y/x+2=k,则直线kx-y+2k=0与圆(x-1)2+(y-2)2=4有交点,所以|k-2+2k|/(k~2+1)~(1/2)≤2 解得0≤k≤12/5,故y/x+2∈[0,12/5]. 例2求函数y=sinx/2-cosx的值域. 解:由原函数式得ycosx+sinx-2y=0. 令u=cosx,v=sinx,则直线yu+v-2y= 0与圆u2+v2=1有交点,所以+-2y|/(y~2+1/~(1/2))≤1.
Constructing a straight line and a circle with an intersection, using the point-line distance formula can be a simple solution to many problems. Example 1 If the real number x, y for the equation x2+y2-2x-4y +1=0. Then take the algebraic formula y/x+2 The value range is ____. Solution: Let y/x+2=k, then the line kx-y+2k=0 and the circle (x-1)2+(y-2)2=4 have intersections, so |k- 2+2k|/(k~2+1)~(1/2)≤2 The solution is 0≤k≤12/5, so y/x+2∈[0,12/5]. Example 2 Find the function y The range of =sinx/2-cosx. Solution: ycosx+sinx-2y=0 from the original function. Let u=cosx,v=sinx, then the straight line yu+v-2y= 0 and the circle u2+v2=1 There is an intersection, so +-2y|/(y~2+1/~(1/2))≤1.