Question

抛物线Y=-1/2x^2上有两点A(X1,Y1),B(X2,Y2),且向量OA·向量OB=0,又向量OM=(0,-2)

Answer 1

1.

OA*OB = 0

故 -1/2(x1)^2*-1/2(x2)^2 + (x1)*(x2) = 0

即 (x1)*(x2) + 4 = 0

而AM // AB的充要条件是 (y2 - y1)*(-x1) = （-2 - y1）*(x2 - x1)

化简即得(x1)*(x2) + 4 = 0,就是上面得到的结论,得证

2.

MA=-2MB

故 (x1, y1+2) = -2(x2, y2+2)

又 (x1)*(x2) + 4 = 0

得 x1 = -2sqrt(2)(就是-2倍根号2的意思) x2 = sqrt(2)(根号2)

A(-2sqrt(2), -4)

B(sqrt(2), -1)

所以直线AB方程可得

AB ： y = 5/6*(sqrt(2))*x - 8/3