曲线C上任一点到点F1(-4,0),F2(4,0)的距离之和为12.曲线C的左顶点为A,点P在曲线C上,且PA⊥PF2.

1个回答

  • (1)依题意,曲线C上是以F1(-4,0),F2(4,0)为焦点的椭圆

    其中,c=4,2a=12,a=6,b^2=20

    曲线C的方程:x^2/36+y^2/20=1

    (2)设 P(x,y) ,A(-6,0),F2(4,0)

    ∵ PA⊥PF2 ,P在以PA为直径的圆上

    ∴ (x+1)^2+y^2=25与x^2/36+y^2/20=1联立

    解得x=-6(舍去) 或x=3/2,

    ∴ P(3/2,±5√3/2)

    (3)设M(0,m),N(x,y),x^2=36(1-y^2/20) -√20≤y≤√20

    (本问的意思:将 M(0,m)看成定点,在曲线C上找

    动点N使MN最大,而这个最大值恰好是 3√7)

    |MN|^2=x^2+(y-m)^2= 36-9/5 y^2+y^2-2my+m^2

    =-4/5 y^2 -2my+m^2+36

    =-4/5(y+5m/4)^2+9/4 m^2+36

    当-√20≤-5m/4≤√20时,-4√20/5 ≤m≤4√20/5

    y=-5m/4,|MN|^2取得最大值9m^2/4+36

    9m^2/4+36=(3√7)^2,m^2=12,m=±2√3

    当- 5m/4>√20,m<-4√20/5

    y=√20 ,|MN|^2取得最大值 20-2√20m+m^2

    20-2√20m+m^2=63,m^2-2√20m-43=0

    解得:m=√20-3√7>-4√20/5 (舍去)

    当- 5m/4<-√20,m>4√20/5

    y=-√20 ,|MN|^2取得最大值 20+2√20m+m^2

    20+2√20m+m^2=63,m^2+2√20m-43=0

    解得:m=-√20+3√7