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苏教版选修2-1习题2.6(3)第10题:一只酒杯的轴截面是抛物线的一部分,它的方程是x2=2y (0≤y≤20).在杯内放入一个玻璃球,要使球触及酒杯底部,那么玻璃球的半径r应满足什么条件?李洪洋在文[1]中给出一种常规解法:设抛物线上任意一点为(x,y),它到点(0,r)的距离为d=(x2+(y-r)2)~(1/2)=(2y+(y-r)2)~(1/2)=(y[-(r-1)]2+2r-1)~(1/2),由题意知,抛物线顶点0,(0)是抛物线上到0,(r)距离最近的点,从而r-1≤0,即r≤1.又
Sujiao version elective 2-1 Exercise 2.6 (3) Question 10: a glass axis of the parabolic section is part of its equation is x2 = 2y (0 ≤ y ≤ 20) in the cup into a glass ball , To make the ball touch the bottom of the glass, then the radius of the glass ball r should meet the conditions? Li Hong Yang [1] gives a conventional solution: Let parabolic point at any point (x, y) , r) is given by the following formula: d = (x2 + (yr) 2) ~ (1/2) = (2y + (yr) 2) ~ (1/2) = (y [- (r- 1)] 2 + 2r- 1) ~ (1/2), from the title, the parabola vertex 0, (0) is the parabola to 0, (r) the nearest point, so r-1 ≤ 0, that is, r ≤ 1.