满足点(A，B)和方程式时的抛物线

给定一个点(A，B) ，使得曲线上每个点的距离My = Nx ² + Ox + P (A，B)等于曲线上该点与x轴之间的距离。任务是找到M，NO和P的值。

注意：该方程式应为简单形式，即gcd(| M |，| N |，| O |，| P |)= 1且N始终为正。

例子：

Input: A = 1, B = 1 
Output: 2 1 -2 2 
M = 2, N = 1, O = -2, P = 2 
The equation of the curve will be 
2y = x² – 2x + 2

Input: A = -1, B = 1 
Output: 2 1 2 2 

方法：从抛物线的属性来看，对于曲线上的每个点，其与准线和焦点的距离将始终相等。使用此属性，将y = 0当作一个准线，将(A，B)当作焦点。
由于方程中N始终为正，因此抛物线将朝上，抛物线的方程为(x – h) ² = 4p(y – k) ，其中(h，k)是顶点和p是焦点和顶点或顶点和方向之间的距离。

由于顶点是形成焦点(A，B)的垂直距离和此处的方向(A，0)之间的中点，因此为B。因此，顶点的坐标为(A，B / 2)，并且p也是B / 2 。
因此，等式为(x – A) ² = 4 * B / 2 *(y – B / 2) 。
可以求解此方程式以得出结果：

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the required values
void solve(int A, int B)
{
    double p = B / 2.0;
    int M = ceil(4 * p);
    int N = 1;
    int O = - 2 * A;
    int Q = ceil(A * A + 4 * p*p);
    cout << M << " " << N << " "
         << O << " " << Q;
}
 
// Driver code
int main()
{
    int a = 1;
    int b = 1;
    solve(a, b);
}
 
// This code is contributed by Mohit Kumar

Java

// Java implementation of the approach
class GFG
{
     
    // Function to find the required values
    static void solve(int A, int B)
    {
        double p = B / 2.0;
        double M = Math.ceil(4 * p);
        int N = 1;
        int O = - 2 * A;
        double Q = Math.ceil(A * A + 4 * p * p);
        System.out.println(M + " " + N +
                               " " + O + " " + Q);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int a = 1;
        int b = 1;
        solve(a, b);
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to find the required values
def solve(A, B):
    p = B / 2
    M = int(4 * p)
    N = 1
    O = - 2 * A
    Q = int(A * A + 4 * p*p)
    return [M, N, O, Q]
 
# Driver code
a = 1
b = 1
print(*solve(a, b))

C#

// C# implementation of the approach
using System;
 
class GFG
{
    // Function to find the required values
    static void solve(int A, int B)
    {
        double p = B / 2.0;
         
        double M = Math.Ceiling(4 * p);
         
        int N = 1;
        int O = - 2 * A;
         
        double Q = Math.Ceiling(A * A + 4 * p * p);
         
        Console.Write(M + " " + N + " " + O + " " + Q);
    }
     
    // Driver code
    static public void Main ()
    {
        int a = 1;
        int b = 1;
        solve(a, b);
    }
}
 
// This code is contributed by AnkitRai01

输出：

2 1 -2 2