直线; 根号2ax + by =1 与圆 x^2+ y^2=1 相交于A ,B (其中a ,b两点是实数)且三角形AOB是直角三角形(O是坐标原点)则点p(a,b)与点(0,1)之间的距离最大值为( )思路是 圆心(0,0)到直线2ax+by-1=0
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![直线; 根号2ax + by =1 与圆 x^2+ y^2=1 相交于A ,B (其中a ,b两点是实数)且三角形AOB是直角三角形(O是坐标原点)则点p(a,b)与点(0,1)之间的距离最大值为( )思路是 圆心(0,0)到直线2ax+by-1=0](/uploads/image/z/12472306-34-6.jpg?t=%E7%9B%B4%E7%BA%BF%3B+%E6%A0%B9%E5%8F%B72ax+%2B+by+%3D1+%E4%B8%8E%E5%9C%86+x%5E2%2B+y%5E2%3D1+%E7%9B%B8%E4%BA%A4%E4%BA%8EA+%2CB+%28%E5%85%B6%E4%B8%ADa+%2Cb%E4%B8%A4%E7%82%B9%E6%98%AF%E5%AE%9E%E6%95%B0%EF%BC%89%E4%B8%94%E4%B8%89%E8%A7%92%E5%BD%A2AOB%E6%98%AF%E7%9B%B4%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2%EF%BC%88O%E6%98%AF%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%EF%BC%89%E5%88%99%E7%82%B9p%EF%BC%88a%2Cb%29%E4%B8%8E%E7%82%B9%EF%BC%880%2C1%EF%BC%89%E4%B9%8B%E9%97%B4%E7%9A%84%E8%B7%9D%E7%A6%BB%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA%EF%BC%88+%EF%BC%89%E6%80%9D%E8%B7%AF%E6%98%AF+%E5%9C%86%E5%BF%83%280%2C0%29%E5%88%B0%E7%9B%B4%E7%BA%BF2ax%EF%BC%8Bby%EF%BC%8D1%3D0)
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直线; 根号2ax + by =1 与圆 x^2+ y^2=1 相交于A ,B (其中a ,b两点是实数)且三角形AOB是直角三角形(O是坐标原点)则点p(a,b)与点(0,1)之间的距离最大值为( )思路是 圆心(0,0)到直线2ax+by-1=0
直线; 根号2ax + by =1 与圆 x^2+ y^2=1 相交于A ,B (其中a ,b两点是实数)且三角形AOB是直角三角形(O是坐标原点)则点p(a,b)与点(0,1)之间的距离最大值为( )
思路是 圆心(0,0)到直线2ax+by-1=0的距离是√2/2,即得:1/√[4a²+b²]=√2/2,得:4a²+b²=2,而d=√[a²+(b-1)²]=√[(3/4)b²-2b+(3/2)]转化为求二次函数最值问题,其中还要考虑-√2≤b√2,d的最大值是当b=√2取得的,是√2-1.想问b的取值范围是怎么求出来的呢?
直线; 根号2ax + by =1 与圆 x^2+ y^2=1 相交于A ,B (其中a ,b两点是实数)且三角形AOB是直角三角形(O是坐标原点)则点p(a,b)与点(0,1)之间的距离最大值为( )思路是 圆心(0,0)到直线2ax+by-1=0
4a²+b²=2
4a^2>=0
b^2