高二向量与椭圆综合题

1)x²/(5m/2)+y²/(3m/2)=1

a²=5m/2,b²=3m/2,∴c²=a²-b²=m

F(√m,0), 而直线AB斜率为1,则AB:y=x-√m

设A(x1,y1),B(x2,y2)

y=x-√m代入椭圆方程 x²/(5m/2)+(x-√m)²/(3m/2)=1

整理得 8x²-10√mx-5m/2=0

∴x1+x2=5√m/4,x1x2=-5m/16

∴y1+y2=(x1-√m)+(x2-√m)=(x1+x2)-2√m=-3√m/4

∴M(5√m/8,-3√m/8)

∴直线OM:y=-3x/5

代入椭圆方程 x²/(5m/2)+(9x²/25)/(3m/2)=1

解得 x=5√m/4,y=-3x/5=-3√m/4, N(5√m/4,-3√m/4)

向量OA+向量OB=(x1,y1)+(x2,y2)=(x1+x2,y1+y2)=(5√m/4,-3√m/4)

向量ON=(5√m/4,-3√m/4)

∴向量OA+向量OB=向量ON

2)y1y2=(x1-√m)(x2-√m)=x1x2-√m(x1+x2)+m=-5m/16-√m×(5√m/4)+m=-9m/16

∴向量OA*向量OB=(x1,y1)(x2,y2)=x1x2+y1y2=-5m/16-9m/16=-7m/8

c=m，点F坐标(m,o),线AB方程：y=x-m，代入椭圆方程

6x^2+10(x-m)^2=15m^2,

16x^2-20mx-5m^2=0,

故M点横坐标为5m/8,

因为x1^2/5+y1^2/3=x2^2/5+y2^2/3，从而3(x1+x2)(x1-x2)=5(y1+y2)(y2-y1), -3·5m/4=5(y1+y2)

M(5m/8,-3m/8)

又直线3x+5y=0交椭圆于N(5m/4, -3m/4),故M为ON中点， OANB为平行四边形

所以OA+OB=ON

OA·OB=x1x2+y1y2=x1x2+9/25 x1x2=34/25 ·(-5m^2/16)=-17m^2/40