Satisfy the parabola when point (A, B) and the equation is given

Given a point (A, B) such that distance from each point on the curve M.y = N.x2 + O.x + P with (A, B) is equal to the distance between that point on the curve and x-axis. The task is to find the value of M, N O and P.
Note: The equtation should be in simple form i.e. gcd( |M|, |N|, |O|, |P| ) = 1 and N should always be positive.

Examples:

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

Approach: From the property of the parabola, for every point on the curve the distance with the directrix and the focus will always be equal. Using this porperty, consider y = 0 as a directrix and (A, B) as the focus.
Since in the equation N is given to be always positive, the parabola will be facing upward and the equation of the parabola is (x – h)2 = 4p(y – k) where (h, k) is the co-ordinate of the vertex and p is the distance between focus and vertex or vertex and directrix.
Since the vertex is the mid point between the perpendicular distance form focus (A, B) and the directrix here (A, 0) is B. So the co-ordinate of the vertex is (A, B/2) and p is also B/2.
So, the equation will be (x – A)2 = 4 * B/2 * (y – B/2).
This equation can be solved to get the result:

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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

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Java

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// 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

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Python3

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# 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))

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C#

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// 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 

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Output:

2 1 -2 2


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Improved By : mohit kumar 29, AnkitRai01

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