若直线ax+y+2=0平分双曲线 x^2/16-y^2/9=1的斜率为1的弦,则a的取值范围是 求详细过程

设双曲线斜率为1的弦所在的直线方程为y=x+by=x+b与 x^2/16-y^2/9=1联立消去y得：x^2/16-（x+b)^2/9=1即 7x^2+32bx+16b^2+144=0Δ=1024b^2-28(16b^2+144)>0==>b√7设弦端点A(x1,y1),B(x2,y2),AB中点M(x',y')x1+x2=-32b/7,x'=-16b/7 ,y'=-9b/7ax+y+2=0过M点∴-16ab/7-9b/7+2=0==>16ab=14-9ba=7/(8b)-9/16b√7∴ -√7/8-9/16