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试题设圆满足：①截y轴所得弦长为ZF＠技工轴分成两段圆弧，其弧长的比为3：1．在满足条件①，②的所有固中，求协。到直线L：X一Zy一0的距离最小的目的方程．解法fib所求的团为（—一d）‘十（y—b）’一厂‘，由①，②易得rZ—a’＋1，／一Zb‘，消去厂得Zb’一a’＝1．可见
The test questions are set to satisfy the circle: 1 y-axis The resulting chord length is ZF@. The mechanic axis is divided into two arcs with a ratio of arc length of 3:1. In all the solids satisfying conditions 1, 2, seek agreement. The objective equation with the smallest distance to the straight line L:X-Zy-0. The group sought by the solution fib is (--d) ’ten (y-b) ’a factory’, from 1,2 is easy to get rZ-a’+1, /a Zb’, eliminate the factory to get Zb’a’ = 1. visible