已知X型的椭圆标准方程，那么如何设与它共焦点的双曲线方程？为什么？
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\u89e3\uff1ac^2\u5df2\u77e5
\u8bbe\u4e3ax^2/a^2-y^2/(c^2-a^2)=1
\u5982\u6709\u7591\u95ee\uff0c\u53ef\u8ffd\u95ee\uff01
焦点为(c,0) (-c,0)
c^2=a^2-b^2
x^2/d^2-y^2/e^2=1
d^2+e^2=c^2
d^2+e^2=a^2-b^2
就共焦点了
绛旓細妞渾鐨勬爣鍑嗘柟绋嬪叡鍒嗕袱绉嶆儏鍐 锛氬綋鐒︾偣鍦x杞存椂锛屾き鍦嗙殑鏍囧噯鏂圭▼鏄細x^2/a^2+y^2/b^2=1锛(a>b>0)锛涘綋鐒︾偣鍦▂杞存椂锛屾き鍦嗙殑鏍囧噯鏂圭▼鏄細y^2/a^2+x^2/b^2=1锛(a>b>0)锛
绛旓細c^2=a^2-b^2 c
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