求过(3,-2)且与x^2/9+y^2/4=1有相同焦点的椭圆方程
问题描述：
求过(3,-2)且与x^2/9+y^2/4=1有相同焦点的椭圆方程
答
x^2/9+y^2/4=1a^2=9,b^2=4c^2=5,设椭圆方程为：x^2/a^2+y^2/b^2=1则：a^2=b^2+5即：x^2/(b^2+5)+y^2/b^2=1把：x=3,y=-2代人得：9/(b^2+5)+4/b^2=1解得：b^2=10a^2=b^2+5=15椭圆方程为：x^2/15+y^2/10=1...