求焦点为F1(0,-6),F2(0,6),且过点M(2,5)的双曲线的标准方程.
问题描述:
求焦点为F1(0,-6),F2(0,6),且过点M(2,5)的双曲线的标准方程.
答
c=6
c^2=a^2+b^2=36
设:双曲线方程为y^2/a^2-x^2/b^2=1
即y^2/a^2-x^2/(36-a^2)=1
因为过点(2,5)
代入
所以
25/a^2-4/(36-a^2)=1
25(36-a^2)-4a^2=a^2(36-a^2)
a^4-65a^2+900=0
(a^2-20)(a^2-45)=0
因为a
b^2=16
所以方程为y^2/20-x^2/16=1
答
设方程是y^2/a^2 - x^2/b^2 = 1
a^2+b^2=36
又因为过点(2,5)
25/a^2 - 4/b^2 = 1
解得
a^2=16
b^2=20
y^2/20-x^2/16=1
答
焦点为F1(0,-6),F2(0,6),
c=6,焦点在y轴
y^2/b^2-x^2/a^2=1
a^2+b^2=c^2
b^2=36-a^2
把M代入
25/(36-a^2)-4/a^2=1
25a^2-4(36-a^2)=a^2(36-a^2)
a^4-7a^2-144=0
(a^2-16)(a^2+9)=0
a^2>0
a^2=16
b^2=20
所以y^2/20-x^2/16=1