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 Area, Amax = 2ab

Below is the implementation of the above approach:

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// 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;
}
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// 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));
}
}
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# 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
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// 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
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<?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
?>
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Output:
160

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