Question

第十题求助数学

Answer 1

设A(xa,a),B(xb,yb),
向量OA*向量OB=xaxb+yayb=(ya^2yb^2)/16+yayb (1式）
又因点A点B与点M在一条直线上，可得ya/(xa-1)=yb/(xb-1),yayb/yb(xa-1）=yayb/ya(xb-1)

化简得yb(xa-1)=ya(xb-1),yb(ya^2/4-1)=ya(yb^2/4-1),
最终得到yayb=-1，带入(1式）得-15/16

Answer 2

A(x1,y1), B(x2,y2)

直线方程，过点(2,0)

y=m(x-2)                        (1)

y^2=4x                           (2)

sub (1) into (2)

m^2.(x-2)^2 = 4x

m^2.x^2 -(4m^2+4)x +4m^2 =0

x1.x2 =4m^2/m^2 =4

Similary

y^2 =4(y/m +2)

my^2 =4(y+2m)

my^2-4y -8m=0

y1.y2 = -8m/m = -8

OA.OB = x1.x2 +y1.y2 = 4-8 =-4

ans : B