Given an ellipse, with major axis length 2a & 2b, the task is to find the area of the largest triangle that can be inscribed in it.
Examples:
Input: a = 4, b = 2 Output: 10.3923 Input: a = 5, b = 3 Output: 10.8253
Approach: So we know the ellipse is just the scaled shadow of a circle.Let’s find the scaling factor.
x^2/a^2 + y^2/b^2 = 1 is an ellipse. Rewrite this as:
(y*(a/b))^2+x^2 = a^2
This is just a vertically scaled down circle of radius a (think light falls from the top at an angle), and the vertical factor is a/b. The biggest triangle in the ellipse is then a scaled up version of the biggest triangle in the circle. Using a little geometry and taking symmetry into account, we can understand that the biggest such triangle is the equilateral one. It’s sides will be ?3a and the area will be (3?3)a^2/4
Translating this to ellipse terms – we scale the horizontal dimension up by a factor a/b, and the area of the biggest triangle in the ellipse is,
A = (3?3)a^2/4b
Below is the implementation of above approach:
C++
// C++ Program to find the biggest triangle// which can be inscribed within the ellipse#include <bits/stdc++.h>using namespace std;// Function to find the area// of the trianglefloat trianglearea(float a, float b){ // a and b cannot be negative if (a < 0 || b < 0) return -1; // area of the triangle float area = (3 * sqrt(3) * pow(a, 2)) / (4 * b); return area;}// Driver codeint main(){ float a = 4, b = 2; cout << trianglearea(a, b) << endl; return 0;} |
Java
//Java Program to find the biggest triangle//which can be inscribed within the ellipsepublic class GFG { //Function to find the area //of the triangle static float trianglearea(float a, float b) { // a and b cannot be negative if (a < 0 || b < 0) return -1; // area of the triangle float area = (float)(3 * Math.sqrt(3) * Math.pow(a, 2)) / (4 * b); return area; } //Driver code public static void main(String[] args) { float a = 4, b = 2; System.out.println(trianglearea(a, b)); }} |
Python3
# Python 3 Program to find the biggest triangle # which can be inscribed within the ellipsefrom math import *# Function to find the area # of the triangledef trianglearea(a, b) : # a and b cannot be negative if a < 0 or b < 0 : return -1 # area of the triangle area = (3 * sqrt(3) * pow(a, 2)) / (4 * b) return area# Driver Codeif __name__ == "__main__" : a, b = 4, 2 print(round(trianglearea(a, b),4))# This code is contributed by ANKITRAI1 |
C#
// C# Program to find the biggest // triangle which can be inscribed// within the ellipseusing System;class GFG{// Function to find the area// of the trianglestatic float trianglearea(float a, float b){// a and b cannot be negativeif (a < 0 || b < 0) return -1;// area of the trianglefloat area = (float)(3 * Math.Sqrt(3) * Math.Pow(a, 2)) / (4 * b);return area;}// Driver codepublic static void Main(){ float a = 4, b = 2; Console.WriteLine(trianglearea(a, b));}}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP Program to find the biggest// triangle which can be inscribed // within the ellipse// Function to find the area// of the trianglefunction trianglearea($a, $b){ // a and b cannot be negative if ($a < 0 || $b < 0) return -1; // area of the triangle $area = (3 * sqrt(3) * pow($a, 2)) / (4 * $b); return $area;}// Driver code$a = 4;$b = 2;echo trianglearea($a, $b); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script>// javascript Program to find the biggest triangle// which can be inscribed within the ellipse// Function to find the area// of the trianglefunction trianglearea(a , b){ // a and b cannot be negative if (a < 0 || b < 0) return -1; // area of the triangle var area = (3 * Math.sqrt(3) * Math.pow(a, 2)) / (4 * b); return area;}// Driver codevar a = 4, b = 2;document.write(trianglearea(a, b).toFixed(4));// This code contributed by shikhasingrajput </script> |
10.3923
Time Complexity: O(1)
Auxiliary Space: O(1)
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