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

  • Last Updated : 10 Jun, 2021
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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 equation 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 property, 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++




// 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

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

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Function to find the required values
    function solve(A, B)
    {
        let p = B / 2.0;
        let M = Math.ceil(4 * p);
        let N = 1;
        let O = - 2 * A;
        let Q = Math.ceil(A * A + 4 * p * p);
        document.write(M + " " + N + " " + O + " " + Q);
    }
     
    let a = 1;
    let b = 1;
    solve(a, b);
     
</script>
Output: 
2 1 -2 2

 

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