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有关弦的中点问题，应灵活运用“点差法”，“设而不求法”来简化运算．
来源学§科§网Z§X§X§K
考点一圆锥曲线的弦长及中点问题例1x2y26已知椭圆G：2＋2＝1ab0的离心率为，右焦点22，0，斜率为1的直线lab3与椭圆G交于A，B两点，以AB为底边作等腰三角形，顶点为P－32．1求椭圆G的方程；2求△PAB的面积．解c61由已知得c＝22，＝a3
解得a＝23，又b2＝a2－c2＝4x2y2所以椭圆G的方程为＋＝11242设直线l的方程为y＝x＋m
y＝x＋m，由x2y212＋4＝1
得4x2＋6mx＋3m2－12＝0①设A，B的坐标分别为x1，y1，x2，y2x1x2，AB中点为Ex0，y0，x1＋x23mm则x0＝＝－，y0＝x0＋m＝；244因为AB是等腰△PAB的底边，所以PE⊥ABm2－4所以PE的斜率k＝＝－13m－3＋4解得m＝2此时方程①为4x2＋12x＝0解得x1＝－3，x2＝0所以y1＝－1，y2＝2所以AB＝32此时，点P－32到直线AB：
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