两道双曲线题(详细点,有分加)1.以动圆与两圆:x²+y²=1和x²+y²-8x+12=0都外切,则动圆心的轨迹为?2.如图,已知梯形ABCD中AB=2CD,点E分有向线段AC所成的比为8/11,双曲线过C、D、E三点,
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![两道双曲线题(详细点,有分加)1.以动圆与两圆:x²+y²=1和x²+y²-8x+12=0都外切,则动圆心的轨迹为?2.如图,已知梯形ABCD中AB=2CD,点E分有向线段AC所成的比为8/11,双曲线过C、D、E三点,](/uploads/image/z/6112102-22-2.jpg?t=%E4%B8%A4%E9%81%93%E5%8F%8C%E6%9B%B2%E7%BA%BF%E9%A2%98%EF%BC%88%E8%AF%A6%E7%BB%86%E7%82%B9%2C%E6%9C%89%E5%88%86%E5%8A%A0%EF%BC%891.%E4%BB%A5%E5%8A%A8%E5%9C%86%E4%B8%8E%E4%B8%A4%E5%9C%86%EF%BC%9Ax%26sup2%3B%2By%26sup2%3B%3D1%E5%92%8Cx%26sup2%3B%2By%26sup2%3B-8x%2B12%3D0%E9%83%BD%E5%A4%96%E5%88%87%2C%E5%88%99%E5%8A%A8%E5%9C%86%E5%BF%83%E7%9A%84%E8%BD%A8%E8%BF%B9%E4%B8%BA%3F2.%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E6%A2%AF%E5%BD%A2ABCD%E4%B8%ADAB%3D2CD%2C%E7%82%B9E%E5%88%86%E6%9C%89%E5%90%91%E7%BA%BF%E6%AE%B5AC%E6%89%80%E6%88%90%E7%9A%84%E6%AF%94%E4%B8%BA8%2F11%2C%E5%8F%8C%E6%9B%B2%E7%BA%BF%E8%BF%87C%E3%80%81D%E3%80%81E%E4%B8%89%E7%82%B9%2C)
两道双曲线题(详细点,有分加)1.以动圆与两圆:x²+y²=1和x²+y²-8x+12=0都外切,则动圆心的轨迹为?2.如图,已知梯形ABCD中AB=2CD,点E分有向线段AC所成的比为8/11,双曲线过C、D、E三点,
两道双曲线题(详细点,有分加)
1.以动圆与两圆:x²+y²=1和x²+y²-8x+12=0都外切,则动圆心的轨迹为?
2.如图,已知梯形ABCD中AB=2CD,点E分有向线段AC所成的比为8/11,双曲线过C、D、E三点,且以A、B为焦点,求双曲线的离心率.
两道双曲线题(详细点,有分加)1.以动圆与两圆:x²+y²=1和x²+y²-8x+12=0都外切,则动圆心的轨迹为?2.如图,已知梯形ABCD中AB=2CD,点E分有向线段AC所成的比为8/11,双曲线过C、D、E三点,
1 两已知圆的圆心和半径分别是C1(0,0),r1=1; C2 (4,0) ,r2=2,
设动圆圆心为M,半径为R,由于与两已知圆都外切,所以:
MC1=r1+R,MC2=r2+R,相减得:MC2 -MC1=r2-r1=1.
有双曲线定义,点M的轨迹是以C1,C2为焦点的双曲线的左支.
2 以AB方向为X轴正向AB垂直平分线为Y轴建立坐标系,设A(-c,0),B(c,0),由于
AB=2CD,所以可设C(c/2,m),D(-c/2,m),再设E(x,y),因为点E分有向线段AC所成的比
为8/11,所以由向量AE=8/19倍向量AC(或由定比分点坐标公式),可得:
19(x+c,y)=8(3c/2,m),解得:x=-7c/19,y=8m/19,即E(-7c/19,8m/19)
设双曲线方程为:x^2/a^2-y^2/b^2=1,把CE两点坐标代入得:
c^2/(4a^2) - m^2/b^2=1,49c^2/(361a^2)-64m^2/361b^2=1,
消去m得:c^2/a^2=9,所以离心率e=c/a=3