题目
Problem
A curve has the equation
y=2x2−53.
Find the equation of the normal to the curve at the point (2,1), giving your answer in the form ax+by+c=0, where a, b and c are integers.
[6]
题目中文翻译
一条曲线的方程为
y=2x2−53
求该曲线在点 (2,1) 处的法线方程，并将答案写成 ax+by+c=0 的形式，其中 a、b 和 c 是整数。
[6]
解答
解法一
思路
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用链式法则求切线斜率。法线与切线垂直，斜率取负倒数，最后整理成整数系数。
答题过程
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For y=3(2x2−5)−1,
y′=−(2x2−5)212x.
At x=2, this is −24/9=−8/3, so the normal slope is 3/8. Its equation is
y−1=83(x−2),3x−8y+2=0.
