Skip to content
Related Articles
Get the best out of our app
GFG App
Open App
geeksforgeeks
Browser
Continue

Related Articles

Program to find the Eccentricity of an Ellipse

Improve Article
Save Article
Like Article
Improve Article
Save Article
Like Article

Given two positive integers A and B, representing the length of the semi-major and semi-minor axis of an ellipse of the equation \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1         , the task is to find the eccentricity of the given ellipse.

Examples:

Input: A = 12, B = 9
Output: 0.66

Input: A = 6, B = 3
Output: 0.87

Approach: The given problem can be solved based on the following formula to calculate the eccentricity of an ellipse which is given by:

\sqrt(1 - \frac{B^2}{A^2})
where, 
A = Length of semi major axis
B = Length of semi minor axis

Therefore, 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 ellipse
void findEccentricity(double A,
                      double B)
{
    // Store the squares of length of
    // semi-major and semi-minor axis
    double semiMajor = A * A;
    double semiMinor = B * B;
 
    // Calculate the eccentricity
    double ans = sqrt(1 - semiMinor / semiMajor);
 
    // Print the result
    cout << fixed << setprecision(2)
         << ans;
}
 
// Driver Code
int main()
{
    double A = 12, B = 9;
    findEccentricity(A, B);
 
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the
// eccentricity of ellipse
static void findEccentricity(double A, double B)
{
    // Store the squares of length of
    // semi-major and semi-minor axis
    double semiMajor = A * A;
    double semiMinor = B * B;
     
    // Calculate the eccentricity
    double ans = Math.sqrt(1 - semiMinor / semiMajor);
 
    // Print the result
    System.out.format("%.2f", ans);
}
 
// Driver Code
public static void main(String arr[])
{
    double A = 12, B = 9;
    findEccentricity(A, B);
}
}
 
// This code is contributed by kirti


Python3




# Python3 program for the above approach
import math
 
# Function to find the
# eccentricity of ellipse
def findEccentricity(A, B):
     
    # Store the squares of length of
    # semi-major and semi-minor axis
    semiMajor = A * A
    semiMinor = B * B
 
    # Calculate the eccentricity
    ans = math.sqrt(1 - semiMinor / semiMajor)
 
    # Print the result
    print('%.2f' % ans)
 
# Driver Code
if __name__ == "__main__":
 
    A = 12
    B = 9
     
    findEccentricity(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 ellipse
static void findEccentricity(double A, double B)
{
     
    // Store the squares of length of
    // semi-major and semi-minor axis
    double semiMajor = A * A;
    double semiMinor = B * B;
      
    // Calculate the eccentricity
    double ans = Math.Sqrt(1 - semiMinor / semiMajor);
  
    // Print the result
    Console.Write(Math.Round(ans, 2));
}
 
// Driver code
static void Main()
{
    double A = 12, B = 9;
     
    findEccentricity(A, B);
}
}
 
// This code is contributed by code_hunt


Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the
// eccentricity of ellipse
function findEccentricity(A, B)
{
      var semiMajor = A * A;
      var semiMinor = B * B;
      var ans = Math.sqrt(1 - semiMinor / semiMajor)
       
      return ans.toFixed(2);
}
  
// Driver Code
var A = 12;
var B = 9;
 
document.write(findEccentricity(A, B));
    
// This code is contributed by bunnyram19
           
</script>


Output: 

0.66

 

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


My Personal Notes arrow_drop_up
Last Updated : 04 May, 2021
Like Article
Save Article
Similar Reads
Related Tutorials