方程(k+1)x^2+(1-k)y^2=16表示两条直线平行,则k的值为?

Question

方程(k+1)x^2+(1-k)y^2=16表示两条直线平行,则k的值为?

Answer 1

L1:ax+by+c[1]=0
L2:ax+by+c[2]=0
c[1]≠c[2]
(ax+by+c[1])(ax+by+c[2])=0
(a^2)x^2+(b^2)y^2+(c[1]+c[2])(ax+by)+2abxy+c[1]c[2]=0
(k+1)x^2+(1-k)y^2=16
(k+1)x^2+(1-k)y^2-16=0
a^2=k+1,b^2=1-k,(c[1]+c[2])(ax+by)=0,2ab=0,c[1]c[2]=-16
由(c[1]+c[2])(ax+by)=0得c[1]+c[2]=0或者a=b=0,又2ab=0,则显然a、b不能同时为0
故,有c[1]+c[2]=0、a=0或b=0
当a=0时,k=-1
当b=0时,k=1
故k为-1或1