曲线y=13x2在点R(8,14)的切线方程是()A.x+48y-20=0B.x+48y+20=0C.x-48y+20=0D.x-4y-20=0-数学

题目简介

曲线y=13x2在点R(8,14)的切线方程是()A.x+48y-20=0B.x+48y+20=0C.x-48y+20=0D.x-4y-20=0-数学

题目详情

曲线y=
1
3x2
在点R(8,
1
4
)的切线方程是(  )
A.x+48y-20=0B.x+48y+20=0C.x-48y+20=0D.x-4y-20=0
题型:单选题难度:中档来源:不详

答案

由y=class="stub"1
3x2
=x-class="stub"2
3
,得到y′=-class="stub"2
3
x-class="stub"5
3

则切线的斜率k=-class="stub"3
2
8-class="stub"5
3
=-class="stub"2
3
(23)-class="stub"5
3
=-class="stub"1
48

所以切线方程是:y-class="stub"1
4
=-class="stub"1
48
(x-8),化简得x+48y-20=0.
故选A.

更多内容推荐