Given two integers A and B, representing the length of the semi-major and semi-minor axes of a Hyperbola, the task is to find the length of the latus rectum of the hyperbola.
Examples:
Input: A = 3, B = 2
Output: 2.66666Input: A = 6, B = 3
Output: 3
Approach: The Latus Rectum of a hyperbola is the focal chord perpendicular to the major axis and the length of the Latus Rectum is equal to (Length of the minor axis )2/(length of major axis).
Follow the steps below to solve the given problem:
- Find the length of the major axis of the hyperbola and store it in a variable, say major.
- Find the length of the minor axis of the hyperbola and store it in a variable, say minor.
- After completing the above steps, print the value of (minor*minor)/major as the resultant length of the Latus Rectum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to calculate the length of// the latus rectum of a hyperboladouble lengthOfLatusRectum(double A, double B){ // Store the length of major axis double major = 2.0 * A; // Store the length of minor axis double minor = 2.0 * B; // Store the length of the // latus rectum double latus_rectum = (minor * minor) / major; // Return the length of the // latus rectum return latus_rectum;}// Driver Codeint main(){ double A = 3.0, B = 2.0; cout << lengthOfLatusRectum(A, B); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to calculate the length of// the latus rectum of a hyperbolastatic double lengthOfLatusRectum(double A, double B){ // Store the length of major axis double major = 2.0 * A; // Store the length of minor axis double minor = 2.0 * B; // Store the length of the // latus rectum double latus_rectum = (minor * minor) / major; // Return the length of the // latus rectum return latus_rectum;}// Driver Codepublic static void main(String[] args){ double A = 3.0, B = 2.0; System.out.println(lengthOfLatusRectum(A, B));}}// This code is contributed by Dharanendra L V. |
Python3
# Python program for the above approach# Function to calculate the length of# the latus rectum of a hyperboladef lengthOfLatusRectum(A,B): # Store the length of major axis major = 2.0 * A # Store the length of minor axis minor = 2.0 * B # Store the length of the # latus rectum latus_rectum = (minor * minor) / major # Return the length of the # latus rectum return latus_rectum# Driver CodeA = 3.0B = 2.0print(round(lengthOfLatusRectum(A, B),5))# This code is contributed by avanitrachhadiya2155 |
C#
// C# program for the above approachusing System;class GFG{// Function to calculate the length of// the latus rectum of a hyperbolastatic double lengthOfLatusRectum(double A, double B){ // Store the length of major axis double major = 2.0 * A; // Store the length of minor axis double minor = 2.0 * B; // Store the length of the // latus rectum double latus_rectum = (minor * minor) / major; // Return the length of the // latus rectum return latus_rectum;}// Driver Codepublic static void Main (){ double A = 3.0, B = 2.0; Console.WriteLine(lengthOfLatusRectum(A, B));}}// This code is contributed by ukasp. |
Javascript
<script>// Javascript program for the above approach // Function to calculate the length of// the latus rectum of a hyperbolafunction lengthOfLatusRectum(A, B){ // Store the length of major axis var major = 2.0 * A; // Store the length of minor axis var minor = 2.0 * B; // Store the length of the // latus rectum var latus_rectum = (minor * minor) / major; // Return the length of the // latus rectum return latus_rectum;}// Driver Codevar A = 3.0, B = 2.0;document.write(lengthOfLatusRectum(A, B));// This code is contributed by 29AjayKumar </script> |
Output:
2.66667
Time Complexity: O(1)
Auxiliary Space: O(1)
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