提问曲线方程问题直线l:y=mx+1交椭圆C:ax^2+y^2=2于A、B两点,以OA、OB为邻边的平行四边形OAPB,当a=2时,求点P轨迹方程.
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![提问曲线方程问题直线l:y=mx+1交椭圆C:ax^2+y^2=2于A、B两点,以OA、OB为邻边的平行四边形OAPB,当a=2时,求点P轨迹方程.](/uploads/image/z/4068186-42-6.jpg?t=%E6%8F%90%E9%97%AE%E6%9B%B2%E7%BA%BF%E6%96%B9%E7%A8%8B%E9%97%AE%E9%A2%98%E7%9B%B4%E7%BA%BFl%3Ay%EF%BC%9Dmx%EF%BC%8B1%E4%BA%A4%E6%A4%AD%E5%9C%86C%EF%BC%9Aax%EF%BC%BE2%EF%BC%8By%EF%BC%BE2%EF%BC%9D2%E4%BA%8EA%E3%80%81B%E4%B8%A4%E7%82%B9%2C%E4%BB%A5OA%E3%80%81OB%E4%B8%BA%E9%82%BB%E8%BE%B9%E7%9A%84%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2OAPB%2C%E5%BD%93a%EF%BC%9D2%E6%97%B6%2C%E6%B1%82%E7%82%B9P%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B.)
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