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
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![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](/uploads/image/z/962746-34-6.jpg?t=%E6%B1%82%E4%BB%A5%E6%A4%AD%E5%9C%86x%5E2%2F16%2By%5E2%2F9%3D1%E7%9F%AD%E8%BD%B4%E7%9A%84%E4%B8%A4%E4%B8%AA%E9%A1%B6%E7%82%B9%E4%B8%BA%E7%84%A6%E7%82%B9%2C%E4%B8%94%E8%BF%87%E7%82%B9A%284%2C-5%29%E7%9A%84%E5%8F%8C%E6%9B%B2%E7%BA%BF%E7%9A%84%E6%A0%87%E5%87%86%E6%96%B9%E7%A8%8B.%E8%A7%A3%E6%9E%90%EF%BC%9A%E6%A4%AD%E5%9C%86%E7%9F%AD%E8%BD%B4%E5%9C%A8Y%E8%BD%B4%2C%E6%95%85%E5%8F%8C%E6%9B%B2%E7%BA%BF%E7%84%A6%E7%82%B9%E5%9C%A8Y%E8%BD%B4%2C%E5%85%B6%E7%84%A6%E7%82%B9%E4%B8%BAF1%280%2C-3%29%2CF2%280%2C3%29%2C%E8%AE%BE%E6%96%B9%E7%A8%8B%E4%B8%BA%3Ay%5E2%2Fm%5E2-x%5E2%2Fn%5E2%3D1%2Cm%5E2%2Bn%5E2%3D9%2Cn%5E2%3D9-m%5E2%2Cy%5E2%2Fm%5E2-x%5E2%2F%289)
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
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?
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
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",
将双曲线方程两边和乘以m^2*(9-m^2)即得(通分)。