分析:由焦点坐标可知,焦点在y轴上,可设方程为 y^2/a^2 -x^2/b^2 =1(a>0,b>0),又知c=6,再把点代入即可求得.

解:设所求的双曲线方程为 y^2/a^2 -x^2/b^2 =1(a>0,b>0),则有{■(25/a^2 "-" 4/b^2 =1"," @a^2+b^2=6^2 "," )┤解得{■(a=2√5 "," @b=4"." )┤

　　故所求的双曲线的标准方程为 y^2/20-x^2/16=1.

★10.已知双曲线的焦点在坐标轴上,且双曲线经过M(1,1),N(-2,5)两点,求双曲线的标准方程.

分析:此题由于不知道焦点在哪个坐标轴上,所以应先分两种情况来讨论,再把两点代入.此题还可以先设双曲线的方程为Ax2+By2=1(AB<0),再把两点代入求解.

解法一当焦点在x轴上时,设所求的双曲线的标准方程为 x^2/a^2 -y^2/b^2 =1(a>0,b>0).因为M(1,1),N(-2,5)两点在双曲线上,

　　所以{■(1/a^2 "-" 1/b^2 =1"," @〖"(-" 2")" 〗^2/a^2 "-" 5^2/b^2 =1"," )┤

　　解得{■(a^2=7/8 "," @b^2=7"." )┤

　　当焦点在y轴上时,设双曲线的标准方程为 y^2/a^2 -x^2/b^2 =1(a>0,b>0),

　　同理,有{■(1/a^2 "-" 1/b^2 =1"," @5^2/a^2 "-" 〖"(-" 2")" 〗^2/b^2 =1"," )┤

　　解得{■(a^2="-" 7"," @b^2="-" 7/8 "," )┤舍去.

　　故所求的双曲线的标准方程为 x^2/(7/8)-y^2/7=1.

解法二设所求的双曲线的标准方程为Ax2+By2=1(AB<0).

　　因为M(1,1),N(-2,5)两点在双曲线上,代入上述方程有{■(A+B=1"," @4A+25B=1"," )┤

　　解得{■(A=8/7 "," @B="-" 1/7 "." )┤

故所求的双曲线的标准方程为 x^2/(7/8)-y^2/7=1.