已知斜率为1的直线l与双曲线C:x²/a²-y²/b²=1(a>0,b>0)相较于B,D两点,且BD的中点为M(1,3)1.求C的离心率2.设C的右顶点为A,右焦点为F,|DF|·|BF|=17,证明:过A、B、D三点的圆与x轴相
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![已知斜率为1的直线l与双曲线C:x²/a²-y²/b²=1(a>0,b>0)相较于B,D两点,且BD的中点为M(1,3)1.求C的离心率2.设C的右顶点为A,右焦点为F,|DF|·|BF|=17,证明:过A、B、D三点的圆与x轴相](/uploads/image/z/11503541-29-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%9C%E7%8E%87%E4%B8%BA1%E7%9A%84%E7%9B%B4%E7%BA%BFl%E4%B8%8E%E5%8F%8C%E6%9B%B2%E7%BA%BFC%EF%BC%9Ax%26%23178%3B%2Fa%26%23178%3B-y%26%23178%3B%2Fb%26%23178%3B%3D1%EF%BC%88a%EF%BC%9E0%2Cb%EF%BC%9E0%EF%BC%89%E7%9B%B8%E8%BE%83%E4%BA%8EB%2CD%E4%B8%A4%E7%82%B9%2C%E4%B8%94BD%E7%9A%84%E4%B8%AD%E7%82%B9%E4%B8%BAM%EF%BC%881%2C3%EF%BC%891.%E6%B1%82C%E7%9A%84%E7%A6%BB%E5%BF%83%E7%8E%872.%E8%AE%BEC%E7%9A%84%E5%8F%B3%E9%A1%B6%E7%82%B9%E4%B8%BAA%2C%E5%8F%B3%E7%84%A6%E7%82%B9%E4%B8%BAF%2C%7CDF%7C%C2%B7%7CBF%7C%3D17%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E8%BF%87A%E3%80%81B%E3%80%81D%E4%B8%89%E7%82%B9%E7%9A%84%E5%9C%86%E4%B8%8Ex%E8%BD%B4%E7%9B%B8)
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已知斜率为1的直线l与双曲线C:x²/a²-y²/b²=1(a>0,b>0)相较于B,D两点,且BD的中点为M(1,3)1.求C的离心率2.设C的右顶点为A,右焦点为F,|DF|·|BF|=17,证明:过A、B、D三点的圆与x轴相
已知斜率为1的直线l与双曲线C:x²/a²-y²/b²=1(a>0,b>0)相较于B,D两点,且BD的中点为M(1,3)
1.求C的离心率
2.设C的右顶点为A,右焦点为F,|DF|·|BF|=17,证明:过A、B、D三点的圆与x轴相切
已知斜率为1的直线l与双曲线C:x²/a²-y²/b²=1(a>0,b>0)相较于B,D两点,且BD的中点为M(1,3)1.求C的离心率2.设C的右顶点为A,右焦点为F,|DF|·|BF|=17,证明:过A、B、D三点的圆与x轴相
1.点差法求斜率代入点坐标可得
2.求圆心坐标是关键