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Program to find the Eccentricity of a Hyperbola
  • Last Updated : 13 May, 2021

Given two integers A and B, representing the length of the semi-major and semi-minor axis of a Hyperbola of the equation (X2 / A2) – (Y2 / B2) = 1, the task is to calculate the eccentricity of the given hyperbola.

Examples:

Input: A = 3, B = 2
Output: 1.20185
Explanation:
The eccentricity of the given hyperbola is 1.20185.

Input: A = 6, B = 3
Output: 1.11803

Approach: The given problem can be solved by using the formula to find the eccentricity of an ellipse.



  • The length of the semi-major axis is A.
  • The length of the semi-minor axis is B.
  • Therefore, the eccentricity of the ellipse is given by \sqrt(1 + \frac{B^2}{A^2})           where A > B

Therefore, the idea is to print the value of \sqrt(1 + \frac{B^2}{A^2})           as the eccentricity of the ellipse.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the eccentricity
// of a hyperbola
double eccHyperbola(double A, double B)
{
    // Stores the squared ratio
    // of major axis to minor axis
    double r = (double)B * B / A * A;
 
    // Increment r by 1
    r += 1;
 
    // Return the square root of r
    return sqrt(r);
}
 
// Driver Code
int main()
{
    double A = 3.0, B = 2.0;
    cout << eccHyperbola(A, B);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the eccentricity
// of a hyperbola
static double eccHyperbola(double A, double B)
{
     
    // Stores the squared ratio
    // of major axis to minor axis
    double r = (double)B * B / A * A;
 
    // Increment r by 1
    r += 1;
 
    // Return the square root of r
    return Math.sqrt(r);
}
 
// Driver Code
public static void main(String[] args)
{
    double A = 3.0, B = 2.0;
     
    System.out.print(eccHyperbola(A, B));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program for the above approach
import math
 
# Function to find the eccentricity
# of a hyperbola
 
 
def eccHyperbola(A, B):
 
    # Stores the squared ratio
    # of major axis to minor axis
    r = B * B / A * A
 
    # Increment r by 1
    r += 1
 
    # Return the square root of r
    return math.sqrt(r)
 
 
# Driver Code
if __name__ == "__main__":
 
    A = 3.0
    B = 2.0
    print(eccHyperbola(A, B))
 
    # This code is contributed by ukasp

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the eccentricity
// of a hyperbola
static double eccHyperbola(double A, double B)
{
     
    // Stores the squared ratio
    // of major axis to minor axis
    double r = (double)B * B / A * A;
 
    // Increment r by 1
    r += 1;
 
    // Return the square root of r
    return Math.Sqrt(r);
}
 
// Driver Code
public static void Main(String[] args)
{
    double A = 3.0, B = 2.0;
     
    Console.Write(eccHyperbola(A, B));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the eccentricity
// of a hyperbola
function eccHyperbola(A, B)
{
     
    // Stores the squared ratio
    // of major axis to minor axis
    let r = B * B / A * A;
 
    // Increment r by 1
    r += 1;
 
    // Return the square root of r
    return Math.sqrt(r);
}
 
// Driver Code
let A = 3.0;
let B = 2.0;
 
document.write(eccHyperbola(A, B));
 
// This code is contributed by mohit kumar
 
</script>
Output: 
2.23607

 

Time Complexity: O(1)
Auxiliary Space: O(1)

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