# 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.x ^{2} + 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 = x^{2}– 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++

`// 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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)) ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

**Output:**

2 1 -2 2

## Recommended Posts:

- Check if a point is inside, outside or on the parabola
- Number of sextuplets (or six values) that satisfy an equation
- Equation of straight line passing through a given point which bisects it into two equal line segments
- Finding the vertex, focus and directrix of a parabola
- Reflection of a point at 180 degree rotation of another point
- Find x, y, z that satisfy 2/n = 1/x + 1/y + 1/z
- Rotation of a point about another point in C++
- Count of sub-sequences which satisfy the given condition
- Pair of integers (a, b) which satisfy the given equations
- Find numbers a and b that satisfy the given conditions
- Pairs from an array that satisfy the given condition
- Count all possible N digit numbers that satisfy the given condition
- Find n positive integers that satisfy the given equations
- Count index pairs which satisfy the given condition
- Count of indices in an array that satisfy the given condition

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.