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Minimum area of the triangle formed by any tangent to an ellipse with the coordinate axes
  • Last Updated : 07 Apr, 2021

Given two integers A and B, representing the length of the semi-major and semi-minor axis of an ellipse with the equation (x2 / A2) + (y2 / B2) = 1, the task is to find the minimum area of the triangle formed by any tangent to the ellipse with the coordinate axes.

Examples:

Input: A = 1, B = 2
Output: 2

Input: A = 2, B = 3
Output: 6

 

Approach:



The idea is based on the observation that the equation of a tangent at coordinate (A * cosθ, B * sinθ) on the ellipse in the figure shown above is:

 (x * cosθ / A) + (y * sinθ / B) = 1

The coordinates of P and Q are (A / cosθ, 0) and (0, B / sinθ) respectively.
Area of triangle formed by the tangent to the ellipse and the coordinate axes is given by:

Area of ΔOPQ = 0.5 * base * height = 0.5 * (A / cosθ) * (B / sinθ) = (A * B) / (2 * sinθ * cosθ) = (A * B) / sin2θ
                                                                                    
Therefore, Area = (A * B) / sin2θ — (1)

To minimize the area, the value of sin2θ should be maximum possible, i.e. 1.
Therefore, the minimum area possible is A * B.

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 minimum area
// of triangle formed by any tangent
// to ellipse with the coordinate axes
void minimumTriangleArea(int a, int b)
{
    // Stores the minimum area
    int area = a * b;
 
    // Print the calculated area
    cout << area;
}
 
// Driver Code
int main()
{
    int a = 1, b = 2;
    minimumTriangleArea(a, b);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
 
// Function to find the minimum area
// of triangle formed by any tangent
// to ellipse with the coordinate axes
static void minimumTriangleArea(int a, int b)
{
     
    // Stores the minimum area
    int area = a * b;
 
    // Print the calculated area
    System.out.println(area);
}
 
// Driver Code
public static void main(String[] args)
{
    int a = 1, b = 2;
     
    minimumTriangleArea(a, b);
}
}
 
// This code is contributed by AnkThon

Python3




# Python3 program for the above approach
 
# Function to find the minimum area
# of triangle formed by any tangent
# to ellipse with the coordinate axes
def minimumTriangleArea(a, b):
     
    # Stores the minimum area
    area = a * b
 
    # Print the calculated area
    print(area)
 
# Driver Code
a = 1
b = 2
 
minimumTriangleArea(a, b)
 
# This code is contributed by rohitsingh07052

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the minimum area
// of triangle formed by any tangent
// to ellipse with the coordinate axes
static void minimumTriangleArea(int a, int b)
{
     
    // Stores the minimum area
    int area = a * b;
 
    // Print the calculated area
    Console.WriteLine(area);
}
 
// Driver Code
public static void Main()
{
    int a = 1, b = 2;
     
    minimumTriangleArea(a, b);
}
}
 
// This code is contributed by ukasp

Javascript




// JavaScript program for the above approach
 
// Function to find the minimum area
// of triangle formed by any tangent
// to ellipse with the coordinate axes
function minimumTriangleArea(a, b)
{
     
    // Stores the minimum area
    var area = a * b
 
    // Print the calculated area
   console.log(area)
     
}
 
// Driver Code
var a = 1
var b = 2
 
minimumTriangleArea(a, b)
 
// This code is contributed by AnkThon
Output
2

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

 

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