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Area of the Largest square that can be inscribed in an ellipse
  • Last Updated : 15 Mar, 2021

Given an ellipse, with major axis length 2a & 2b, the task is to find the area of the largest rectangle that can be inscribed in it.
Examples: 
 

Input: a = 4, b = 2
Output: 1.25

Input: a = 5, b= 3
Output: 0.604444

 

 

Approach: If a square is inscribed in an ellipse, the distance from the centre of the square to any of its corners will be equal to the distance between the origin and the point on the upper right corner in the diagram below, where x=y
 



the equation of the ellipse is x^2/a^2 + y^2/b^2 = 1 
If, x = y 
then, x^2/a^2 + x^2/b^2 = 1 
therefore, x = √(a^2 + b^2)/ab 
so, y = √(a^2 + b^2)/ab 
So Area, A = 4(a^2 + b^2)/a^2b^2
 

Below is the implementation of above approach:
 

C++




// C++ Program to find the biggest square
// which can be inscribed within the ellipse
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the area
// of the square
float squarearea(float a, float b)
{
 
    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;
 
    // area of the square
    float area = 4 * ((pow(a, 2) + pow(b, 2))
                      / (pow(a, 2) * pow(b, 2)));
 
    return area;
}
 
// Driver code
int main()
{
    float a = 4, b = 2;
    cout << squarearea(a, b) << endl;
 
    return 0;
}

Java




// Java Program to find the biggest square
// which can be inscribed within the ellipse
import java.io.*;
 
class GFG {
 
 
// Function to find the area
// of the square
static float squarearea(float a, float b)
{
 
    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;
 
    // area of the square
    float area = 4 *(float) ((Math.pow(a, 2) + Math.pow(b, 2))
                    / (Math.pow(a, 2) * Math.pow(b, 2)));
 
    return area;
}
 
// Driver code
 
    public static void main (String[] args) {
        float a = 4, b = 2;
    System.out.println( squarearea(a, b));
 
    }
}
// This code is contributed by inder_verma.

Python 3




# Python3 Program to find the biggest square
# which can be inscribed within the ellipse
 
 
# Function to find the area
# of the square
def squarearea( a, b):
 
 
    # a and b cannot be negative
    if (a < 0 or b < 0):
        return -1
     
 
    # area of the square
    area = 4 * (((pow(a, 2) + pow(b, 2)) /
               (pow(a, 2) * pow(b, 2))))
 
    return area
 
 
# Driver code
if __name__=='__main__':
    a = 4
    b = 2
    print(squarearea(a, b))
 
# This code is contributed by ash264

C#




// C# Program to find the biggest
// square which can be inscribed
// within the ellipse
using System;
 
class GFG
{
 
// Function to find the area
// of the square
static float squarearea(float a, float b)
{
 
    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;
 
    // area of the square
    float area = 4 *(float) ((Math.Pow(a, 2) +
                              Math.Pow(b, 2)) /
                             (Math.Pow(a, 2) *
                              Math.Pow(b, 2)));
 
    return area;
}
 
// Driver code
public static void Main ()
{
    float a = 4, b = 2;
    Console.WriteLine( squarearea(a, b));
}
}
 
// This code is contributed by inder_verma

PHP




<?php
// PHP Program to find the biggest square
// which can be inscribed within the ellipse
 
 
// Function to find the area
// of the square
function squarearea( $a, $b)
{
 
    // a and b cannot be negative
    if ($a < 0 or $b < 0)
        return -1;
     
 
    // area of the square
    $area = 4 * (((pow($a, 2) + pow($b, 2)) /
            (pow($a, 2) * pow($b, 2))));
 
    return $area;
}
 
// Driver code
 
    $a = 4;
    $b = 2;
    print(squarearea($a, $b));
 
// This code is contributed by mits
?>

Javascript




<script>
 
// javascript Program to find the biggest square
// which can be inscribed within the ellipse
 
// Function to find the area
// of the square
function squarearea(a , b)
{
 
    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;
 
    // area of the square
    var area = 4 *((Math.pow(a, 2) + Math.pow(b, 2))
                    / (Math.pow(a, 2) * Math.pow(b, 2)));
 
    return area;
}
 
// Driver code
var a = 4, b = 2;
document.write( squarearea(a, b));
 
// This code contributed by shikhasingrajput
 
</script>
Output: 
1.25

 

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