一道高二数学圆锥曲线的题已知三角形P1OP2的面积为27/4,向量P1P=2向量PP2,求以直线OP1,OP2为渐近线且

e=根号13/2
=>
b^2/a^2=9/4
=>两渐进线为
y=1.5x
y=-1.5x
设
|OP1|=a |OP2|=b
2*S三角形=absin∠P1OP2
而tan(∠P1OP2/2)=2/3
sin∠P1OP2=12/13
ab=117/2
a=|x1|*根号13/2
所以
|x1*x2|=18
设P(x3,y3)
p1p/pp2=2,
=>
x3=2x2-x1
y3=2y2-y1
(2x2-x1)^2/4-(2y2-y1)^2/9=k
=>
-2x1*x2=k
k=36
x^2/144-y^/324=1