题目

已知F1、F2是双曲线x^2-y^2=1的两个焦点
点M是双曲线右支上的点,O为坐标原点,若|MF1|+|MF2|/|MO|=根号6,点M的坐标为

提问时间：2020-11-14

答案

F1(-√2,0) ,F2(√2,0),设M(x,y)
则：x^2-y^2=1 |MO|= √(x^2+y^2) |MF1|=√[(x+√2)^2+y^2] |MF2|=√[(x-√2)^2+y^2]
√[(x+√2)^2+y^2]+√[(x-√2)^2+y^2]/√(x^2+y^2)=√6
解得M的坐标：
（-√2/2{1+[3√6+27+3√(6+18√6)]^(1/3)/3+3/[3√6+27+3√(6+18√6)]^(1/3)+2/3*√6},
1/6/(3√6+27+3√(6+18√6))^(2/3)*√2*((3√6+27+3√(6+18√6))^(2/3)*(18√(6+18√6)+33*(3√6+27+3√(6+18√6))^(2/3)+126√6+315+12√6√(6+18√6)+54*(3√6+27+3√(6+18√6))^(1/3)+12√6*(3√6+27+3√(6+18√6))^(2/3)+(3√6+27+3√(6+18√6))^(4/3)+36√6*(3√6+27+3√(6+18√6))^(1/3)))^(1/2)）
或
（ -√2/2{1+[3√6+27+3√(6+18*√6))^(1/3)/3+3/[3√6+27+3√(6+18√6))^(1/3)+2/3*√6)},
-1/6/(3√6+27+3√(6+18√6))^(2/3)*√2*((3√6+27+3√(6+18√6))^(2/3)*(18√(6+18√6)+33(3√6+27+3√(6+18√6))^(2/3)+126√6+315+12√6√(6+18√6)+54*(3√6+27+3√(6+18√6)^(1/3)+12√6*(3√6+27+3√(6+18√6))^(2/3)+(3√6+27+3√(6+18√6))^(4/3)+36√6*(3√6+27+3√(6+18√6))^(1/3)))^(1/2))