(1)由点a到椭圆c两点的距离之和为4有2a=4
=>
a=2e=c/a=√2/2
=>
c=√2所以b^2=a^2
-
c^2=2
=>
b=√2所以椭圆c的方程为x^2/4
+
y^2/2=1(2)假设存在，则有：因为直线y=2x斜率为k0=2所以直线pp1斜率为k=
-1/2则有
(y1-y0)/(x1-x0)=
-1/2··············①pp1中点坐标为m(
(x1+x0)/2
,
(y1+y0)/2
)因为m在y=2x上所以有
(y1+y0)/2
=
x1+x0················②因为p在椭圆上所以x0^2/4
+
y0^2/2=1······················③另外3x1-4y1=10································④将①②③④联立建立方程组得：若无解，则假设不成立，不存在点p若有解，则p点的坐标为(x0,y0)计算出的结果
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