Area of Largest rectangle that can be inscribed in an Ellipse

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 = 3
Output: 24

Input: a = 10, b = 8
Output: 160

Approach: 
Let the upper right corner of the rectangle has co-ordinates (x, y), 
Then the area of rectangle, A = 4*x*y.
Now, 

Equation of ellipse, (x²/a²) + (y²/b²) = 1
Thinking of the area as a function of x, we have 
dA/dx = 4xdy/dx + 4y
Differentiating equation of ellipse with respect to x, we have 
2x/a² + (2y/b²)dy/dx = 0,
so, 
dy/dx = -b²x/a²y, 
and 
dAdx = 4y – (4b²x²/a²y)
Setting this to 0 and simplifying, we have y² = b²x²/a².
From equation of ellipse we know that, 
y²=b² – b²x²/a2
Thus, y²=b² – y², 2y²=b², and y²b² = 1/2. 
Clearly, then, x²a² = 1/2 as well, and the area is maximized when 
x= a/?2 and y=b/?2
So the maximum area, A_max = 2ab

Below is the implementation of the above approach:  

C++

// C++ Program to find the biggest rectangle
// which can be inscribed within the ellipse
#include <bits/stdc++.h>
using namespace std;

// Function to find the area
// of the rectangle
float rectanglearea(float a, float b)
{

    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;

    // area of the rectangle
    return 2 * a * b;
}

// Driver code
int main()
{
    float a = 10, b = 8;
    cout << rectanglearea(a, b) << endl;
    return 0;
}

Java

// Java Program to find the biggest rectangle
// which can be inscribed within the ellipse

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG{
// Function to find the area
// of the rectangle
static float rectanglearea(float a, float b)
{

    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;

    // area of the rectangle
    return 2 * a * b;
}

// Driver code
public static void main(String args[])
{
    float a = 10, b = 8;
    System.out.println(rectanglearea(a, b));
}
}

Python 3

# Python 3 Program to find the biggest rectangle
# which can be inscribed within the ellipse

#  Function to find the area
# of the rectangle
def rectanglearea(a, b) :

    # a and b cannot be negative
    if a < 0 or b < 0 :
        return -1

    # area of the rectangle
    return 2 * a * b


# Driver code
if __name__ == "__main__" :

    a, b = 10, 8
    print(rectanglearea(a, b))


# This code is contributed by ANKITRAI1

C#

// C# Program to find the
// biggest rectangle which
// can be inscribed within
// the ellipse
using System;

class GFG
{
// Function to find the area
// of the rectangle
static float rectanglearea(float a,
                           float b)
{

    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;

    // area of the rectangle
    return 2 * a * b;
}

// Driver code
public static void Main()
{
    float a = 10, b = 8;
    Console.WriteLine(rectanglearea(a, b));
}
}

// This code is contributed
// by inder_verma

PHP

<?php
// PHP Program to find the biggest
// rectangle which can be inscribed
// within the ellipse

// Function to find the area
// of the rectangle
function rectanglearea($a, $b)
{

    // a and b cannot be negative
    if ($a < 0 or $b < 0)
        return -1;

    // area of the rectangle
    return 2 * $a * $b;
}

// Driver code
$a = 10; $b = 8;
echo rectanglearea($a, $b);

// This code is contributed
// by inder_verma
?>

Javascript

<script>

// javascript Program to find the biggest rectangle
// which can be inscribed within the ellipse

// Function to find the area
// of the rectangle
function rectanglearea(a , b)
{

    // a and b cannot be negative
    if (a < 0 || b < 0)
        return -1;

    // area of the rectangle
    return 2 * a * b;
}

// Driver code

var a = 10, b = 8;
document.write(rectanglearea(a, b));

// This code contributed by Princi Singh

</script>

Output

160

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