点M(x,y)到定点F(0,√7)的距离和它到直线y=(4√7)/7的距离
点M(x,y)到定点F(0,√7)的距离和它到直线y=(4√7)/7的距离的比是常数√7/2,求点M的轨迹方程?
人气:262 ℃ 时间:2020-05-13 12:25:51
解答
点M(x,y)到定点F(0,√7)的距离a=√(x^2+(y-√7))
点M(x,y)到直线y=(4√7)/7的距离b=|y-(4√7)/7|
∵a:b=√7/2
∴√(x^2+(y-√7)^2):|y-(4√7)/7|=√7/2
两边同时平方得:[x^2+(y-√7)^2]/[y-(4√7)/7]^2=7/4
化简得:4x^2-3y^2+12=0
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