一道解析几何题

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已知双曲线,已知向量a=(x/5,y/2√6),向量b=(x/5,-y/2√6),曲线a*b=1上一点M到F(7,0)的距离为11，O为坐标原点，N是MF的中点，求ON的长度

a*b=1
x/5*x/5-y/2√6*y/2√6=1
x^2/25-y^2/24=1..............1

(x-7)^2+y^2=11^2
y^2=121-(x-7)^2.............2

将2式代入1式得
x^2/25-y^2/24=1
x^2/25-[121-(x-7)^2]/24=1
x^2/25-[72-x^2+14x]/24=1
24x^2-25[72-x^2+14x]=600
49x^2-350x-2400=0
(7x-80)(7x+30)=0
x=80/7 或 x=-30/7(舍去)...............3

将3式代入2式得
y^2=121-(x-7)^2
=121-(80/7-7)^2
=121-(31/7)^2
=121-961/49
=4968/49

∣ON∣=√{[(x+7)/2]^2+(y/2)^2}
=√{[(80/7+7)/2]^2+y^2/4}
=√[(129/14)]^2+(4968/49)/4]
=√[16641/196+4968/196]
=√[16641/196+4968/196]
=√(21609/196)
=147/14
=21/2

所以ON=21/2