# Minimum area of the triangle formed by any tangent to an ellipse with the coordinate axes

Given two integers **A** and **B**, representing the length of the semi-major and semi-minor axis of an ellipse with the equation **(x ^{2} / A^{2}) + (y^{2} / B^{2}) = 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 = 2Output:2

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

**Approach:**

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

(x * cosÎ¸ / A) + (y * sinÎ¸ / B) = 1

The coordinates of **P** and **Q** are **(A / cosÎ¸, 0)** and **(0, B / sinÎ¸)** respectively.

Area of triangle formed by the tangent to the ellipse and the coordinate axes is given by:

Area of Î”OPQ = 0.5 * base * height = 0.5 * (A / cosÎ¸) * (B / sinÎ¸) = (A * B) / (2 * sinÎ¸ * cosÎ¸) = (A * B) / sin2Î¸

Therefore, Area = (A * B) / sin2Î¸ â€” (1)

To minimize the area, the value of **sin2Î¸** 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)