设椭圆的离心率=1/2,右焦点F(c,0)方程ax^2+bx-c=0的两个实根为x1,x2.则P(x1,x20必在圆x^2+y^2=2上?设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax^2+bx-c=0的两个实根分别为x1和x2,则点P(x1,
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![设椭圆的离心率=1/2,右焦点F(c,0)方程ax^2+bx-c=0的两个实根为x1,x2.则P(x1,x20必在圆x^2+y^2=2上?设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax^2+bx-c=0的两个实根分别为x1和x2,则点P(x1,](/uploads/image/z/1633164-60-4.jpg?t=%E8%AE%BE%E6%A4%AD%E5%9C%86%E7%9A%84%E7%A6%BB%E5%BF%83%E7%8E%87%3D1%2F2%2C%E5%8F%B3%E7%84%A6%E7%82%B9F%EF%BC%88c%2C0%29%E6%96%B9%E7%A8%8Bax%5E2%2Bbx-c%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9%E4%B8%BAx1%2Cx2.%E5%88%99P%EF%BC%88x1%2Cx20%E5%BF%85%E5%9C%A8%E5%9C%86x%5E2%2By%5E2%3D2%E4%B8%8A%3F%E8%AE%BE%E6%A4%AD%E5%9C%86x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2%3D1%28a%3Eb%3E0%29%E7%9A%84%E7%A6%BB%E5%BF%83%E7%8E%87e%3D1%2F2%2C%E5%8F%B3%E7%84%A6%E7%82%B9F%EF%BC%88c%2C0%29%2C%E6%96%B9%E7%A8%8Bax%5E2%2Bbx-c%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9%E5%88%86%E5%88%AB%E4%B8%BAx1%E5%92%8Cx2%2C%E5%88%99%E7%82%B9P%EF%BC%88x1%2C)
设椭圆的离心率=1/2,右焦点F(c,0)方程ax^2+bx-c=0的两个实根为x1,x2.则P(x1,x20必在圆x^2+y^2=2上?设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax^2+bx-c=0的两个实根分别为x1和x2,则点P(x1,
设椭圆的离心率=1/2,右焦点F(c,0)方程ax^2+bx-c=0的两个实根为x1,x2.则P(x1,x20必在圆x^2+y^2=2上?
设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax^2+bx-c=0的两个实根分别为x1和x2,则点P(x1,x2)()
A.必在圆x^2+y^2=2 内 B.必在圆x^2+y^2=2上
C.必在圆x^2+y^2=2外 D.以上三种情况都有可能
e=根(1-b2/a2)这个是个公式么我怎么没见过能帮我证明下么
设椭圆的离心率=1/2,右焦点F(c,0)方程ax^2+bx-c=0的两个实根为x1,x2.则P(x1,x20必在圆x^2+y^2=2上?设椭圆x^2/a^2+y^2/b^2=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax^2+bx-c=0的两个实根分别为x1和x2,则点P(x1,
由e= ca=12,知 ba=32,由x1,x2是方程ax2+bx-c=0的两个实根,知 x1+x2=-ba=-32, x1x2=-ca=-12,所以x12+x22=(x1+x2)2-2x1x2= 34+1=74<3,由此知点P(x1,x2)必在圆x2+y2=3内.∵e= ca=12,∴ ba=32,
∵x1,x2是方程ax2+bx-c=0的两个实根,
∴由韦达定理: x1+x2=-ba=-32, x1x2=-ca=-12,
所以x12+x22=(x1+x2)2-2x1x2
= 34+1=74<3,
所以点P(x1,x2)必在圆x2+y2=3内.
故选A.