题目

求以椭圆x^2/16+y^2/9=1短轴的两个顶点为焦点,且过点A(4,-5)的双曲线的标准方程.
解析：
椭圆短轴在Y轴,故双曲线焦点在Y轴,其焦点为F1(0,-3),F2(0,3),
设方程为:y^2/m^2-x^2/n^2=1,
m^2+n^2=9,n^2=9-m^2,
y^2/m^2-x^2/(9-m^2)=1,
A(4,-5)是双曲线上一点,代入方程,25/m^2-16/(9-m^2)=1,
m^4-50m^2+225=0,
m^2=45(不合题意,>9),m^2=5,m=√5,
n^2=9-5=4,
双曲线方程为:y^2/5-x^2/4=1.
“m^4-50m^2+225=0",这一步是怎么来的?

提问时间：2020-05-08

答案

A(4,-5)是双曲线上一点,代入得(-5)^2/m^2-4^2/(9-m^2)=1,
去分母两边同时乘以m^2(9-m^2)化简整理得“m^4-50m^2+225=0",