圆锥曲线中的四点共圆

2022-08-18 13:43 作者:因恋相思久 0人读过 | 我要投稿

在双曲线 $%5Cfrac%7Bx%5E2%20%7D%7B2%7D-%20%5Cfrac%7By%5E2%20%7D%7B2%7D%20%3D1$ ， A为左顶点，直线 $l%EF%BC%9Ax%3Dt%EF%BC%880%3Ct%3C1%EF%BC%89$ 交x轴于T,过点T的直线 $l_%7B1%7D%20$ 交双曲线于MN两点，连接AM，AN交直线 $l$ 于G，H两点，问A,O,G,H能否在同一个圆上，若能请求出T坐标，若不能，请说明具体理由

若A，O，G，H共圆，那么 $%E2%88%A0GAT%3D%E2%88%A0OHG$ ,即 $%E2%96%B3ATG%E2%88%BD%E2%96%B3HYO$

有 $%5Cvert%20AT%20%5Cvert%20%5Ccdot%20%5Cvert%20OT%20%5Cvert%3D%20%5Cvert%20GT%20%5Cvert%20%5Ccdot%20%5Cvert%20HT%20%5Cvert%20$

设 $T%EF%BC%88t%EF%BC%8C0%EF%BC%89$ $%EF%BC%8CM%EF%BC%88x_%7B1%7D%EF%BC%8C%20y_%7B1%7D%20%EF%BC%89%EF%BC%8CN%EF%BC%88x_%7B2%7D%EF%BC%8Cy%20_%7B2%7D%20%EF%BC%89%EF%BC%8CA%EF%BC%88-%5Csqrt%7B2%7D%20%2C0%EF%BC%89$

直线AM: $y%3D%5Cfrac%7By_%7B1%7D%20%7D%7Bx_%7B1%7D%2B%5Csqrt%7B2%7D%20%20%7D%20%EF%BC%88x%2B%5Csqrt%7B2%7D%20%EF%BC%89$ ，令 $x%3Dt$ ，得 $y_%7BG%7D%20%3D%5Cfrac%7By_%7B1%7D%EF%BC%88t%2B%5Csqrt%7B2%7D%20%EF%BC%89%20%7D%7Bx_%7B1%7D%2B%5Csqrt%7B2%7D%20%20%7D$

同理得 $y_%7BH%7D%20%3D%5Cfrac%7By_%7B2%7D%EF%BC%88t%2B%5Csqrt%7B2%7D%20%EF%BC%89%20%7D%7Bx_%7B2%7D%2B%5Csqrt%7B2%7D%20%20%7D$

$%20%5Cvert%20GT%20%5Cvert%20%5Ccdot%20%5Cvert%20HT%20%5Cvert%20%3D%5Cvert%20%5Cfrac%7By_%7B1%7Dy_%7B2%7D%20%EF%BC%88t%2B%5Csqrt%7B2%7D%20%EF%BC%89%5E2%20%7D%7B(x_%7B1%7D%2B%5Csqrt%7B2%7D%20)%20(x_%7B2%7D%2B%5Csqrt%7B2%7D%20%20)%7D%5Cvert%20$

$%5Cvert%20AT%20%5Cvert%20%5Ccdot%20%5Cvert%20OT%20%5Cvert%3D%20%5Cvert%20t(t%2B%5Csqrt%7B2%7D%20)%5Cvert%20$

所以 $t%3D%5Cvert%20%5Cfrac%7By_%7B1%7Dy_%7B2%7D%20%EF%BC%88t%2B%5Csqrt%7B2%7D%20%EF%BC%89%20%7D%7B(x_%7B1%7D%2B%5Csqrt%7B2%7D%20)%20(x_%7B2%7D%2B%5Csqrt%7B2%7D%20%20)%7D%5Cvert%20$ ①

设直线MN： $n%EF%BC%88x%2B%5Csqrt%7B2%7D%20%EF%BC%89%2Bmy%3D1$

下面进行齐次化

$x%5E2-%20y%5E2%20%3D2$

$%EF%BC%88x%2B%5Csqrt%7B2%7D-%20%5Csqrt%7B2%7D%20%EF%BC%89%5E2-%20y%5E2%20%3D2$

$%EF%BC%88x%2B%5Csqrt%7B2%7D%EF%BC%89%5E2-2%5Csqrt%7B2%7D%20%EF%BC%88x%2B2%EF%BC%89-%20y%5E2%20%3D0$

$%EF%BC%88x%2B%5Csqrt%7B2%7D%EF%BC%89%5E2-2%5Csqrt%7B2%7D%20%EF%BC%88x%2B2%EF%BC%89%5Bn%EF%BC%88x%2B%5Csqrt%7B2%7D%20%EF%BC%89%2Bmy%5D-%20y%5E2%20%3D0$

$%EF%BC%881-2%5Csqrt%7B2%7Dn%20%EF%BC%89%EF%BC%88x%2B%5Csqrt%7B2%7D%EF%BC%89%5E2-2%5Csqrt%7B2%7D%20m%EF%BC%88x%2B2%EF%BC%89y-%20y%5E2%20%3D0$

$%EF%BC%881-2%5Csqrt%7B2%7Dn%20%EF%BC%89-2%5Csqrt%7B2%7D%20m%EF%BC%88%5Cfrac%7By%7D%7Bx%2B%5Csqrt%7B2%7D%20%7D%20%EF%BC%89-%20%EF%BC%88%5Cfrac%7By%7D%7Bx%2B%0A%5Csqrt%7B2%7D%20%7D%20%EF%BC%89%5E2%20%3D0$

$%20%E6%89%80%E4%BB%A5%5Cfrac%7By_%7B1%7Dy_%7B2%7D%20%20%7D%7B(x_%7B1%7D%2B%5Csqrt%7B2%7D%20)%20(x_%7B2%7D%2B%5Csqrt%7B2%7D%20%20)%7D%3D2%5Csqrt%7B2%7D%20n-1$ ②

又因为直线MN过T

所以 $n%3D%5Cfrac%7B1%7D%7Bt%2B%5Csqrt%7B2%7D%20%7D%20$ ③

联合①②③解得 $t%3D%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D%20$ 故 $T%E7%9A%84%E5%9D%90%E6%A0%87%E4%B8%BA%EF%BC%88%5Cfrac%7B%5Csqrt%7B2%7D%20%7D%7B2%7D%20%2C0%EF%BC%89$

但如果把A，O，H，G轨迹画出来，也可能是条双曲线

有人会联想到2021新高考一卷数学21题

个人认为

其实如果在 $T(%5Cfrac%7B1%7D%7B2%7D%20%2C0)$ 时，是与本题一致的，A，B，P，Q也是满足四点共圆的，但一般情况下新高考21题做法就与本题不同了，本题是 $AT%E2%8A%A5GT$ 恒成立，新高考是只有在 $T(%5Cfrac%7B1%7D%7B2%7D%20%2C0)$ 才满足 $AB%E2%8A%A5PQ$

个人的新高考做法：

（1）略， $x%5E2%20-%5Cfrac%7By%5E2%7D%7B16%7D%20%3D1(x%5Cgeq%201)$

设 $A%EF%BC%88x_%7B1%7D%20%EF%BC%8Cy_%7B1%7D%20%EF%BC%89%EF%BC%8CB%EF%BC%88x_%7B2%7D%20%EF%BC%8Cy_%7B2%7D%20%EF%BC%89%EF%BC%8CT%EF%BC%88%5Cfrac%7B1%7D%7B2%7D%20%EF%BC%8Ct%EF%BC%89%E7%9B%B4%E7%BA%BFAB%EF%BC%8CCD%E6%96%9C%E7%8E%87%E5%88%86%E5%88%AB%E4%B8%BAk_%7B1%7D%EF%BC%8C%20k_%7B2%7D%20$

因为 $%5Cvert%20TA%20%5Cvert%20%5Ccdot%20%5Cvert%20TB%20%5Cvert%20%3D%EF%BC%881%2Bk_%7B1%7D%5E2%EF%BC%89%EF%BC%88x_%7B1%7D-%5Cfrac%7B1%7D%7B2%7D%20%20%EF%BC%89%EF%BC%88x_%7B2%7D-%5Cfrac%7B1%7D%7B2%7D%EF%BC%89$

直线 $AB%3Ay%3Dk_%7B1%7D%20%EF%BC%88x-%5Cfrac%7B1%7D%7B2%7D%20%EF%BC%89%2Bt$ 与 $16x%5E2-%20y%5E2%3D16$ 联立有

$%EF%BC%8816-k_%7B1%7D%5E2%20%EF%BC%89%EF%BC%88x-%5Cfrac%7B1%7D%7B2%7D%20%20%EF%BC%89%5E2%2B%EF%BC%8816-2tk_%7B1%7D%20%EF%BC%89%EF%BC%88x-%5Cfrac%7B1%7D%7B2%7D%20%20%EF%BC%89-%EF%BC%88t%5E2%2B12%20%EF%BC%89%3D0$

$%EF%BC%88x_%7B1%7D-%5Cfrac%7B1%7D%7B2%7D%20%20%EF%BC%89%EF%BC%88x_%7B2%7D-%5Cfrac%7B1%7D%7B2%7D%EF%BC%89%3D%5Cfrac%7Bt%5E2%2B12%7D%7Bk_%7B1%7D%5E2-16%20%7D%20$

$%5Cvert%20TA%20%5Cvert%20%5Ccdot%20%5Cvert%20TB%20%5Cvert%20%3D%5Cfrac%7B%EF%BC%881%2Bk_%7B1%7D%5E2%EF%BC%89%EF%BC%88t%5E2%2B12%EF%BC%89%7D%7Bk_%7B1%7D%5E2-16%7D$ 同理得

$%5Cvert%20TP%20%5Cvert%20%5Ccdot%20%5Cvert%20TQ%20%5Cvert%20%3D%5Cfrac%7B%EF%BC%881%2Bk_%7B2%7D%5E2%EF%BC%89%EF%BC%88t%5E2%2B12%EF%BC%89%7D%7Bk_%7B2%7D%5E2-16%7D$

则 $%5Cfrac%7B1%2Bk_%7B1%7D%5E2%7D%7Bk_%7B1%7D%5E2-16%7D%3D%5Cfrac%7B1%2Bk_%7B2%7D%5E2%7D%7Bk_%7B2%7D%5E2-16%7D$

$1%2B%5Cfrac%7B17%7D%7Bk_%7B1%7D%5E2-16%20%7D%20%3D1%2B%5Cfrac%7B17%7D%7Bk_%7B2%7D%5E2-16%7D%20$

即 $%EF%BC%88k_%7B2%7D%5E2%20-16%EF%BC%89%3D%EF%BC%88k_%7B1%7D%5E2%20-16%EF%BC%89$

$%EF%BC%88k_%7B1%7D%20%2Bk_%7B2%7D%20%EF%BC%89%EF%BC%88k_%7B1%7D%20-k_%7B2%7D%20%EF%BC%89%3D0$

所以 $k_%7B1%7D%3D%20k_%7B2%7D%20%EF%BC%88%E8%88%8D%EF%BC%89%EF%BC%8Ck_%7B1%7D%20%2Bk_%7B2%7D%20%3D0$

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圆锥曲线中的四点共圆

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