与椭圆x2/9+y2/4=1共焦点，且过M(3,-2)的椭圆方程 已知一椭圆与椭圆x²/9+y²/4=1有...
\u8fd9\u9898\uff1a\u4e0e\u692d\u5706x2/9+y2/4=1\u5171\u7126\u70b9\uff0c\u4e14\u8fc7M(3,-2)\u7684\u692d\u5706\u65b9\u7a0b\u89e3\uff1a
\u8bbe\u5171\u7126\u70b9\u7684\u692d\u5706\u65b9\u7a0b\u662fx^2/(9+t)+y^2/(4+t)=1,(t>-4)
(3,-2)\u4ee3\u5165\u5f97\uff1a9/(9+t)+4/(4+t)=1
9(4+t)+4(9+t)=(9+t)(4+t)
72+13t=36+13t+t^2
t^2=36
t=+6\u6216\uff0d6(\u4e0d\u7b26\u5408\uff0c\u820d\uff09
\u6545\u65b9\u7a0b\u662f\uff1ax^2/15+y^2/10=1
\u4e0ex²/a²+y²/b²=1\u6709\u76f8\u540c\u7126\u70b9\u7684\u692d\u5706\u53ef\u4ee5\u8bbe\u4e3a\uff1ax²/(a² + t)+y²/(b² + t) =1
\u6240\u4ee5\uff0c\u8bbe\u8be5\u692d\u5706\u4e3ax²/(9 + t)+y²/(4+ t) =1
\u4ee3\u5165x=3\uff0cy=-2\uff0c\u5f97\uff1a9/\uff089+t\uff09+4/\uff084+t\uff09=1
36+4t+36+9t=36+13t+t²
t1=6
t2=-6\uff08y²\u7684\u7cfb\u6570\u4e3a\u8d1f\uff0c\u820d\u53bb\uff09
\u6240\u4ee5\uff0c\u692d\u5706\u65b9\u7a0b\u4e3a\uff1ax²/15+y²/10=1
(3,-2)代入得：9/(9+t)+4/(4+t)=1
9(4+t)+4(9+t)=(9+t)(4+t)
72+13t=36+13t+t^2
t^2=36
t=+6或－6(不符合，舍）
故方程是：x^2/15+y^2/10=1
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