# Equation of tangent line to a curve at a given point

## Equation of tangent line to a curve at a given point

Definition :-The tangent is a straight line which just touches the curve at a given point. To calculate the equations of these lines we shall make use of the fact that the equation of a straight line passing through the point with coordinates (x1, y1) and having gradient m is given by

Consider a function f(x) such as that shown in figure 1. When we calculate the derivative, f ′ (x), of the function at a point x = a say, we are finding the gradient of the tangent to the graph of that function at that point. Given figure shows the tangent drawn at x = a. The gradient of this tangent is f ′ (a).

F ′ (a) is the gradient of the tangent drawn at x = a.
The equation of the tangent that passes through a point (x1, y1) and has gradient m is given by

Example 1 :-Find the equation of the tangent to the curveat the point where x = 4.

Solution :-first we need to get the equation of the curve into the form we can differentiate.

When x = 4,

Y-coordinate when x = 4,

4(y – 2) = 21(x – 4)
4y – 8 = 21x – 84
4y = 21x – 76
21x – 4y – 76 = 0
Therefore, the equation of the tangent is 21x – 4y – 76 = 0

You can also use this method to find the point of contact of a tangent to a curve when given the equation of the curve and the gradient of the tangent.

Example 2 :-Find the point of contact between the tangent and curve with equation

Solution :-First of all, we need to differentiate the equation of the curve to get an expression for the gradient. Now, we would usually 'sub-in' the value to get the gradient, but this time we don't know what x is. We do know, however that the answer is 4/3 , so we can set this equal to the differentiated expression and solve the equation to find x.

Now we can find the y-coordinate by substituting this into the original equation.