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文[1]给出了下列命题及证明,需指出的是其证明欠妥.下面给出正确的证明. 命题1 A为椭圆内一点,F1为左焦点,F2为右焦点,P为椭圆上任一点,当P为F1A与椭圆的交点时,|PA|+|PF2|最小. 分析:原证明是这样的,在椭圆上任取一点P＇,连接P＇A、P＇F1、P＇F2,以F1为圆心, |AF1|为半径画圆弧交P＇F1于B, 则 |AF1|=|BF1|, 在△AF1P＇中, 因为|P＇A|+|AF1|≥|P＇F1|, 所以|P＇A|≥|P＇F1|-|AF1|=
The following propositions and proofs are given in [1]. It should be pointed out that the proofs are incorrect. The following is the correct proof. Proposition 1 A is the point in the ellipse, F1 is the left focus, F2 is the right focus, and P is the ellipse. One point, when P is the intersection of F1A and the ellipse, |PA|+|PF2|minimum. Analysis: Originally proved to be such, take any point P’ on the ellipse, connect P’A, P’F1, P’F2, With F1 as the center, |AF1| draws a circle for the radius P’F1 to B, then |AF1|=|BF1|, in △AF1P’ because |P’A|+|AF1|≥|P’F1 |, ||P’A|≥|P’F1|-|AF1|=