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: 2Input: 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++ 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 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 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# 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 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 |
2
Time Complexity: O(1)
Auxiliary Space: O(1)