作业帮一道圆锥曲线题求解
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/09/30 09:53:56
一道圆锥曲线题求解
(1)设M(X,Y)
则 kAM=Y/(X+3)
kBM=Y/(X-3)
由题意知：Y/(X+3)*Y/(X-3)=1/3
y^2/(x^2-9)=1/3
x^2-9=3y^2
x^2-3y^2=9
x^2/9-y^2/3=1
故M的方程为x^2/9-y^2/3=1
(2)设直线方程为y=x-2√3
代入
x^2-3(x-2√3)^2=9
x^2-3x^2+12√3x-36=9
-2x^2+12√3x-45=0
2x^2-12√3x+45=0
x1+x2=6√3
x1x2=45/2
|DE|=√{(1+k²)[(x1+x2)²-4x1x2]}
=√(1+1)√[(6√3)^2-4*45/2]
=√2*√18
=6
则 kAM=Y/(X+3)
kBM=Y/(X-3)
由题意知：Y/(X+3)*Y/(X-3)=1/3
y^2/(x^2-9)=1/3
x^2-9=3y^2
x^2-3y^2=9
x^2/9-y^2/3=1
故M的方程为x^2/9-y^2/3=1
(2)设直线方程为y=x-2√3
代入
x^2-3(x-2√3)^2=9
x^2-3x^2+12√3x-36=9
-2x^2+12√3x-45=0
2x^2-12√3x+45=0
x1+x2=6√3
x1x2=45/2
|DE|=√{(1+k²)[(x1+x2)²-4x1x2]}
=√(1+1)√[(6√3)^2-4*45/2]
=√2*√18
=6