已知菱形ABCD的顶点A，C在椭圆x2+3y2=4上，对角线BD所在直线的斜率为1。
（1）当直线BD过点（0，1）时，求直线AC的方程；
（2）当∠ABC=60°时，求菱形ABCD面积的最大值。
解：（1）由题意得直线BD的方程为
因为四边形
为菱形，
所以
于是可设直线AC的方程为
由
得
因为
在椭圆上，
所以
，解得
设
两点坐标分别为
，
则
，
，
，
所以
所以AC的中点坐标为
由四边形
为菱形可知，点
在直线
上，
所以
，解得
所以直线AC的方程为
，即
。
（2）因为四边形
为菱形，且
，
所以
所以菱形
的面积
由（1）可得
，
所以
所以当
时，菱形
的面积取得最大值
。
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