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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++
#include <bits/stdc++.h>
using namespace std;
void minimumTriangleArea(int a, int b)
{
int area = a * b;
cout << area;
}
int main()
{
int a = 1, b = 2;
minimumTriangleArea(a, b);
return 0;
}
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Java
class GFG{
static void minimumTriangleArea(int a, int b)
{
int area = a * b;
System.out.println(area);
}
public static void main(String[] args)
{
int a = 1, b = 2;
minimumTriangleArea(a, b);
}
}
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Python3
def minimumTriangleArea(a, b):
area = a * b
print(area)
a = 1
b = 2
minimumTriangleArea(a, b)
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C#
using System;
class GFG{
static void minimumTriangleArea(int a, int b)
{
int area = a * b;
Console.WriteLine(area);
}
public static void Main()
{
int a = 1, b = 2;
minimumTriangleArea(a, b);
}
}
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Javascript
function minimumTriangleArea(a, b)
{
var area = a * b
console.log(area)
}
var a = 1
var b = 2
minimumTriangleArea(a, b)
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Time Complexity: O(1)
Auxiliary Space: O(1)
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