一道高中数学题第二问的答案我算出来是2/13,可是答案是13/2,所以我想确认一下已知a=(x,0)b=(1,y),(a+√3b)⊥(a-√3b)(1)求点P(x,y)的轨迹C的方程（2）若直线l：y=3x+m(m≠0)与曲线C交于A,B两点,D（0,1）且有|AD|=|BD|,试求m的值
2019-05-23
一道高中数学题第二问的答案我算出来是2/13,可是答案是13/2,所以我想确认一下
已知a=(x,0)b=(1,y),(a+√3b)⊥(a-√3b)(1)求点P(x,y)的轨迹C的方程（2）若直线l：y=3x+m(m≠0)与曲线C交于A,B两点,D（0,1）且有|AD|=|BD|,试求m的值
优质解答
(1)(a+√3b)*(a-√3b)=a^2-3b^2=x^2-3-3y^2=0,点P(x,y)的轨迹方程为x^2/3-y^2=1.
(2)联立直线l与曲线C的方程得：26x^2+18mx+3m^2+3=0,xA+xB=-9m/13.
yA+yB=3(xA+xB)+2m=-27m/13+2m=-m/13.
|AD|=|BD|,则|AD|^2=|BD|^2,xA^2+(yA-1)^2=xB^2+(yB-1)^2.
整理得：xA^2-xB^2=2(yA-yB)-(yA^2-yB^2).(xA+xB)(xA-xB)=2(yA-yB)-(xA+xB)(yA-yB).
-(9m/13)(xA-xB)=2(yA-yB)+(m/13)(yA-yB).-9m/13=2[(yA-yB)/(xA-xB)+(m/13)[(yA-yB)/(xA-xB)]
-9m/13=6+3m/13,12m/13=-6,m=-13/2.
(1)(a+√3b)*(a-√3b)=a^2-3b^2=x^2-3-3y^2=0,点P(x,y)的轨迹方程为x^2/3-y^2=1.
(2)联立直线l与曲线C的方程得：26x^2+18mx+3m^2+3=0,xA+xB=-9m/13.
yA+yB=3(xA+xB)+2m=-27m/13+2m=-m/13.
|AD|=|BD|,则|AD|^2=|BD|^2,xA^2+(yA-1)^2=xB^2+(yB-1)^2.
整理得：xA^2-xB^2=2(yA-yB)-(yA^2-yB^2).(xA+xB)(xA-xB)=2(yA-yB)-(xA+xB)(yA-yB).
-(9m/13)(xA-xB)=2(yA-yB)+(m/13)(yA-yB).-9m/13=2[(yA-yB)/(xA-xB)+(m/13)[(yA-yB)/(xA-xB)]
-9m/13=6+3m/13,12m/13=-6,m=-13/2.