题目内容
答案:(9,12)
解析:=(1,2), =(4,5),∴=(1,2)+2(4,5)=(9,12).∴点P坐标为(9,12).
已知点O(0,0),A(1,-2),动点P满足|PA|=3|PO|,则P点的轨迹方程是
A.8x2+8y2+2x-4y-5=0
B.8x2+8y2-2x-4y-5=0
C.8x2+8y2+2x+4y-5=0
D.8x2+8y2-2x+4y-5=0
A.8x2+8y2+2x-4y-5=0
B.8x2+8y2-2x-4y-5=0
C.8x2+8y2+2x+4y-5=0
D.8x2+8y2-2x+4y-5=0
已知点O(0,0),A(0,b),B(a,a3).若△OAB为直角三角形,则必有( )
A.b=a3
B.b=a3+
C.(b-a3)(b-a3-)=0
D.|b-a3|+=0