题目如图,Rt△OABRt\triangle OABRt△OAB的顶点A(−2,4)A\left(-2,4\right)A(−2,4)在抛物线y=ax2y=ax^{2}y=ax2上,将Rt△OABRt\triangle OABRt△OAB绕点OOO顺时针旋转90∘90^{\circ}90∘,得到△OCD\triangle OCD△OCD,边CDCDCD与该抛物线交于点PPP,则点PPP的坐标为( )A.(2(\sqrt{2}(2,2)\sqrt{2})2)B.(2,2)(2,2)(2,2)C.(2(\sqrt{2}(2,2)2)2)D.(2(2(2,2)\sqrt{2})2)知识点:二次函数的应用章节:函数 / 二次函数