(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