Find Tangent at a given point on the curve

• Last Updated : 14 Jun, 2022

Given a curve [ y = x(A â€“ x) ], the task is to find tangent at given point (x, y) on that curve, where A, x, y are integers.
Examples:

Input: A = 2, x = 2, y = 0
Output: y = -2x - 4
Since y = x(2 - x)
y = 2x - x^2 differentiate it with respect to x
dy/dx = 2 - 2x  put x = 2, y = 0 in this equation
dy/dx = 2 - 2* 2 = -2
equation  => (Y - 0 ) = ((-2))*( Y - 2)
=> y = -2x -4

Input: A = 3, x = 4, y = 5
Output: Not possible
Point is not on that curve

Approach:

1. First find if the given point is on that curve or not.
2. If the point is on that curve then, Find the derivative
3. Calculate the gradient of the tangent by Putting x, y in dy/dx.
4. Determine the equation of the tangent by substituting the gradient of the tangent and the coordinates of the given point into the gradient-point form of the straight line equation, where Equation of normal is Y â€“ y = ( dy/dx ) * (X â€“ x).

Below is the implementation of the above approach:

C++

 // C++ program for find Tangent// on a curve at given point   #include using namespace std;   // function for find Tangentvoid findTangent(int A, int x, int y){    // differentiate given equation    int dif = A - x * 2;       // check that point on the curve or not    if (y == (2 * x - x * x)) {           // if differentiate is negative        if (dif < 0)            cout << "y = "                 << dif << "x" << (x * dif) + (y);           else if (dif > 0)               // differentiate is positive            cout << "y = "                 << dif << "x+" << -x * dif + y;           // differentiate is zero        else            cout << "Not possible";    }}   // Driver codeint main(){    // declare variable    int A = 2, x = 2, y = 0;       // call function findTangent    findTangent(A, x, y);       return 0;}

Java

 // Java program for find Tangent// on a curve at given pointimport java.util.*;import java.lang.*;import java.io.*;   class GFG{    // function for find Tangentstatic void findTangent(int A, int x, int y){    // differentiate given equation    int dif = A - x * 2;        // check that point on the curve or not    if (y == (2 * x - x * x)) {            // if differentiate is negative        if (dif < 0)            System.out.println( "y = "                 + dif + "x" + (x * dif + y));            else if (dif > 0)                // differentiate is positive            System.out.println( "y = "                 + dif + "x+" + -x * dif + y);            // differentiate is zero        else            System.out.println("Not possible");    }}    // Driver codepublic static void main(String args[]){    // declare variable    int A = 2, x = 2, y = 0;        // call function findTangent    findTangent(A, x, y);    } }

Python3

 # Python3 program for find Tangent# on a curve at given point   # function for find Tangentdef findTangent(A, x, y) :       #  differentiate given equation    dif = A - x * 2       #  check that point on the curve or not    if y == (2 * x - x * x) :           # if differentiate is negative        if dif < 0 :               print("y =",dif,"x",(x * dif) + (y))           # differentiate is positive        elif dif > 0 :               print("y =",dif,"x+",-x * dif + y)           # differentiate is zero        else :                           print("Not Possible")                  # Driver code    if __name__ == "__main__" :       # declare variable    A, x, y = 2, 2, 0       # call function findTangent    findTangent(A, x, y)                    # This code is contributed by# ANKITRAI1

C#

 // C# program for find Tangent// on a curve at given point    using System;class GFG{      // function for find Tangentstatic void findTangent(int A, int x, int y){    // differentiate given equation    int dif = A - x * 2;          // check that point on the curve or not    if (y == (2 * x - x * x)) {              // if differentiate is negative        if (dif < 0)            Console.Write( "y = "                 + dif + "x" + (x * dif + y)+"\n");              else if (dif > 0)                  // differentiate is positive            Console.Write( "y = "                 + dif + "x+" + -x * dif + y+"\n");              // differentiate is zero        else            Console.Write("Not possible"+"\n");    }}      // Driver codepublic static void Main(){    // declare variable    int A = 2, x = 2, y = 0;          // call function findTangent    findTangent(A, x, y);      }  }

PHP

 0)               // differentiate is positive            echo "y = ",                \$dif , "x+" , -\$x * \$dif + \$y;           // differentiate is zero        else            echo "Not possible";    }}   // Driver code   // declare variable\$A = 2;\$x = 2;\$y = 0;   // call function findTangentfindTangent(\$A, \$x, \$y);   // This code is contributed by Sachin?>

Javascript



Output:

y = -2x-4

Time Complexity : O(1) ,as we are not using any loop.

Auxiliary Space : O(1) ,as we are not using any extra space.

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