已知双曲线C:X^2-Y^2=1和直线l:y=kx-1,若L与C交于A,B两点,o为原点,三角形AOB面积为根号2,求K
2个回答

y=kx-1

x^2-y^2=1

x^2-(kx-1)^2=1

(1-k^2)x^2+2kx-2=0

x1+x2=2k/(k^2-1)

x1x2=2/(k^2-1)

1/2AB*h

kx-y-1=0

1/(k^2+1)^1/2

(x1-x2)^2+(y1-y2)^2=(1+k^2)(x1-x2)^2=(1+k^2)((x1+x2)^2-4x1x2)=(1+k^2)(4k^2/(k^2-1)^2-8(k^2-1)

1/2ABh=2^1/2

1/4AB^2h^2=2

AB^2h^2=8

1/(k^2+1)*(1+k^2)(4k^2/(k^2-1)^2-8(k^2-1))=8

4k^2/(k^2-1)^2-8(k^2-1)=8

k^2/(k^2-1)^2-2(k^2-1)=2

k^2-2(k^2-1)^3=2