- 高二解析几何题1.已知直线L1;ax
- 1.已知直线L1;ax-2y-2a+4=0,L2:2x+a^2y-2a^2-4=0,其中0
- 1.两直线与两坐标轴的截距是x1=2(a-2)/a,y1=2-a,x2=a^2+2,y2=2(a^2+2)/a^2,围成一四边形,0,a<2,x1<0,y1>0,,x2>0,y2>0,S四边形=1/2[|x2y2|+|x1y1|=a^2+4/a^2+a+4/a,,S最小时a=√2,
直线L1;√2x-2y-2√2=0,L2:x+y-4=0.
2.直线L:y=kx,A(0,-1),B(8,0)关于L的对称点A'(x1,ax1^2),B'(x2,ax2^2),AA',BB'中点在L上:
ax1^2-1=kx1....(1)
ax2^2=k(x2+8)...(2)且
ax1^2+1=-x1/k....(3)
ax2^2=-(x2-8)/k...(4)
解之得k=-2,a=-15/16,L:y=-2x,f(x)=-15/16x^2