椭圆C:x²/a²+y²/b²=1(a>b>0)的左右焦点分别为F1,F2,若点P(x,y)到直线y=kx-1的最大距离为2√2,则k=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/06 18:27:28
![椭圆C:x²/a²+y²/b²=1(a>b>0)的左右焦点分别为F1,F2,若点P(x,y)到直线y=kx-1的最大距离为2√2,则k=?](/uploads/image/z/3813407-71-7.jpg?t=%E6%A4%AD%E5%9C%86C%EF%BC%9Ax%26%23178%3B%2Fa%26%23178%3B%2By%26%23178%3B%2Fb%26%23178%3B%3D1%EF%BC%88a%3Eb%3E0%EF%BC%89%E7%9A%84%E5%B7%A6%E5%8F%B3%E7%84%A6%E7%82%B9%E5%88%86%E5%88%AB%E4%B8%BAF1%2CF2%2C%E8%8B%A5%E7%82%B9P%EF%BC%88x%2Cy%29%E5%88%B0%E7%9B%B4%E7%BA%BFy%3Dkx-1%E7%9A%84%E6%9C%80%E5%A4%A7%E8%B7%9D%E7%A6%BB%E4%B8%BA2%E2%88%9A2%2C%E5%88%99k%3D%3F)
x){d9m̪PS64OڕP~5|#.Zn_eϛv>h{ڱɎ]n:nF:/+t*5vlx>{]+m+t
Zixds/
dcӎٶ6IE4q`J_`g38ig3ֿhd-
Oz_,_1醍@<_;<;Px f2{
椭圆C:x²/a²+y²/b²=1(a>b>0)的左右焦点分别为F1,F2,若点P(x,y)到直线y=kx-1的最大距离为2√2,则k=?
椭圆C:x²/a²+y²/b²=1(a>b>0)的左右焦点分别为F1,F2,
若点P(x,y)到直线y=kx-1的最大距离为2√2,则k=?
椭圆C:x²/a²+y²/b²=1(a>b>0)的左右焦点分别为F1,F2,若点P(x,y)到直线y=kx-1的最大距离为2√2,则k=?
做是能做出来,不过太麻烦,你看看你的题完整吗?
[1+√(k²a²+b²)]/√(k²+1)=2√2再解出k就是答案