Question

高中数学选修1-1的题

已知两点B(6,0)和C（-6，0），设点A与B，C的连线AB，AC的斜率分别为k1,k2,如果k1k2=-4/9,求点A所在曲线的方程，并说明他是何种曲线。

Answer 1

设A(x,y)则
k1*k2=y/(x+6)*y/(x-6)=-4/9
整理得 4x^2+9y^2=144
即x^2/36+y^2/16=1
曲线为椭圆

Answer 2

设A(x,y)则
k1*k2=y/(x+6)*y/(x-6)=-4/9
y^2/x^2-36=-4/9
4x^2-144=-9y^2
4x^2+9y^2=144
x^2/36+y^2/16=1
so 曲线为椭圆

Answer 3

设A(x,y)则
k1*k2=y/(x+6)*y/(x-6)=-4/9
整理得 4x^2+9y^2=144
即x^2/36+y^2/16=1
曲线为椭圆

Answer 4

椭圆。x^2/36+y^2/16=1

Answer 5

设A(x,y)则
k1乘以k2=y/(x+6)乘以y/(x-6)=-4/9
整理得 4x^2+9y^2=144
即x^2/36+y^2/16=1
所以曲线为椭圆