在C ++中找到曲线上给定点的法线
假设我们有一条类似y=x(A-x)的曲线，我们必须找到该曲线上给定点(x,y)的法线。这里A是整数，x和y也是整数。
为了解决这个问题，我们检查给定点是否在曲线上，如果是，则找到该曲线的微分，因此将是-
$$\frac{\text{d}y}{\text{d}x}=A-2x$$
然后将x和y放入dy/dx，然后使用该方程式找到法线-
$$Yy=-\lgroup\frac{\text{d}x}{\text{d}y}\rgroup*\lgroupXx\rgroup$$
示例
#include<iostream>
using namespace std;
void getNormal(int A, int x, int y) {
int differentiation = A - x * 2;
if (y == (2 * x - x * x)) {
if (differentiation < 0)
cout << 0 - differentiation << "y = " << "x" << (0 - x) + (y * differentiation);
else if (differentiation > 0)
cout << differentiation << "y = " << "-x+" << x + differentiation * y;
else
cout << "x = " << x;
}
else
cout << "Not possible";
}
int main() {
int A = 5, x = 2, y = 0;
cout << "Equation of normal is: ";
getNormal(A, x, y);
}输出结果
Equation of normal is: 1y = -x+2