设椭圆x²/a²+y²/b²=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax²+bx-c=0的两个实根分别为x₁、x₂,请证明P(x₁,x₂)必在圆x²+y²=2内.
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![设椭圆x²/a²+y²/b²=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax²+bx-c=0的两个实根分别为x₁、x₂,请证明P(x₁,x₂)必在圆x²+y²=2内.](/uploads/image/z/2486010-66-0.jpg?t=%E8%AE%BE%E6%A4%AD%E5%9C%86x%26%23178%3B%2Fa%26%23178%3B%2By%26%23178%3B%2Fb%26%23178%3B%EF%BC%9D1%28a%EF%BC%9Eb%EF%BC%9E0%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%28c%2C0%29%2C%E6%96%B9%E7%A8%8Bax%26%23178%3B%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%BAx%26%238321%3B%E3%80%81x%26%238322%3B%2C%E8%AF%B7%E8%AF%81%E6%98%8EP%28x%26%238321%3B%2Cx%26%238322%3B%29%E5%BF%85%E5%9C%A8%E5%9C%86x%26%23178%3B%2By%26%23178%3B%3D2%E5%86%85.)
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设椭圆x²/a²+y²/b²=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax²+bx-c=0的两个实根分别为x₁、x₂,请证明P(x₁,x₂)必在圆x²+y²=2内.
设椭圆x²/a²+y²/b²=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax²+bx-c=0的两个实根分别为x₁、x₂,请证明P(x₁,x₂)必在圆x²+y²=2内.
设椭圆x²/a²+y²/b²=1(a>b>0)的离心率e=1/2,右焦点F(c,0),方程ax²+bx-c=0的两个实根分别为x₁、x₂,请证明P(x₁,x₂)必在圆x²+y²=2内.
解析:
由题意可得:a=2c,b=√3*c,其中c>0
则方程ax²+bx-c=0可写为:
2cx²+√3*cx-c=0
即2x²+√3*x-1=0
已知方程的两个实根分别是:x₁、x₂
则由韦达定理可得:
x₁+x₂=-(√3)/2,x₁*x₂=-1/2
所以:x₁²+x₂²
=(x₁+x₂)²-2x₁*x₂
=3/4 +1
=7/4
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