急求已知椭圆方程为x^2*9+y^2/4=1,在椭圆上是否存在点P(x,y)到定点A(a,0))(其中0

P(x,y)在椭圆上,则x^2/9+y^2/4=1,∴y²=4(1-x²/9) (-3≤x≤3)∴|PA|²=(x-a)²+y²=x²-2ax+a²+4(1-x²/9)=(5/9)x²-2ax+a²+4=5/9(x²-18/5ax)+a²+4=5/9(x-9/5...(-3≤x≤3)这个x的取值范围是怎么求的啊？x^2/9+y^2/4=1,==>-3≤x≤3,-2≤y≤2