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人类文明的每一点进步都离不开勇敢者的探索,探索是数学发生、发展的灵魂．引例求过抛物线y2=2px(p>0)的焦点的弦长的最小值．解法设AB的中点为E,分别过A、E、B作准线l的垂线,垂足为D、H、C,由抛物线定义知,|AD|=|AF|, |BC|=|BF|．∴|AB|=|AF|+|BF| =|AD|+|BC|=2|EH|．由图可得|HE|≥|GF|,当且仅当AB与x轴垂直时|HE|=|GF|．即|AB|min=2|GF|=2p．联想由解法得焦点弦AB与x轴垂直时取最小值．抛物线是这样,椭圆、双曲线焦点弦最小值是否也是这样呢?
Every progress of human civilization cannot be separated from the exploration of the brave. Exploration is the soul of the occurrence and development of mathematics. The example seeks the minimum of the chord length of the focus of the parabola y2=2px (p>0). Solution AB is set to the middle point of E, respectively, A, E, B as the vertical line l, vertical foot is D, H, C, defined by the parabolic definition, |AD|=|AF|, |BC|=| BF|. ∴|AB|=|AF|+|BF| =|AD|+|BC|=2|EH|. From the graph, |HE|≥|GF| can be obtained if and only if AB is perpendicular to the x-axis |HE|=|GF|. That is, |AB|min=2|GF|=2p. Lenovo obtains the minimum value when the focus string AB is perpendicular to the x-axis. This is the parabola. Is this also the case for elliptic and hyperbolic focal string minimums?