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求答案!!! 急!!!

来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/06/11 15:50:31
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求答案!!! 急!!!
解题思路: (I)由已知可求,a=2,由点(1, √3/2)在该椭圆上,代入可求b,从而可求椭圆的方程 (II)AB为直径的圆过原点⇔ OA•OB=0⇔x1x2+y1y2=0,从而考虑设直线方程,联立直线于椭圆方程进行求解即可.
解题过程:
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最终答案:略