由具有坐标轴的椭圆的任何切线形成的三角形的最小面积

给定两个整数A和B ，分别表示方程(x ² / A ² )+(y ² / B ² )= 1的椭圆的半长轴和半短轴的长度，任务是找到坐标轴与椭圆的任何切线所形成的三角形的最小面积。

例子：

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

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

方法：

该想法基于以下观察结果：上图中椭圆上的坐标(A *cosθ，B *sinθ)处的切线方程为：

 (x * cosθ / A) + (y * sinθ / B) = 1

P和Q的坐标分别为(A /cosθ，0)和(0，B /sinθ) 。
由椭圆的切线和坐标轴形成的三角形的面积由下式给出：

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)

为了使面积最小， sin2θ的值应为最大可能值，即1。
因此，最小可能面积为A * B。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
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

输出

2

时间复杂度： O(1)
辅助空间： O(1)