Question

双曲线离心率根3,且过点（2,3）,求方程

Answer 1

设方程为：x^2/a^2-y^2/b^2=1,c/a=√3,a^2+b^2=c^2,b^2=2a^2,(2,3)代入无解,故焦点在Y轴,

设方程为：y^2/b^2-x^2/a^2=1,b/c=√3,a^2=2b^2,9/b^2-4/(2b^2)=1,b^2=7,a^2=14,

故方程为：y^2/7-x^2/14=1.