# Tangents and Normals

Before attempting this revision your must have knowledge of what tangents are. How to find gradients of tangents and be able to form general formulas for tangents. This is just an expansion from Gradient of a Tangent to a curve and without any knowledge of tangents be ready for huge headaches.
You know that a tangent to a curve is a straight line which touches the curve at a certain point. Well…
The normal to a curve is the straight line perpendicular to the tangent at the same point as the tangent attachment to the curve. In the following illustration the normal to the curve is the blue line while the orange line is the tangent and the red line the curve.
If the gradient of the tangent is m
…then the gradient of the normal is -1/m
For instance;
if the gradient of a tangent is 4 then the gradient of the normal is -1/4
It is as simple as that, this is how you find the normal to the curve at a certain point but you first need to

**differentiate to find the gradient first.**In most cases when working with tangents and normals there will be a need to find the equations of the tangents and normals at a certain point. In the following examples we shall be exploring just that. To find the equations of either the tangent of the normal we need to know the gradient of the curve and the point of the line then we use the following equation; where m is the gradient and (x1, y1) is the point given.### Finding the equation of the tangent

The following is an example of how to find the equation of a tangent.## Example 1

Find the equation of the tangent to the curve; y = 3x^{2}+ 4 at the point where**x = 1**. First we need to find the gradient**m**by differentiating.**6x**is the general gradient formula but you can use other ways you’re used to, to find the gradient. To find the gradient at**x = 1**we substitute the**x**value in our gradient formula**6x**. So the gradient is a**6**, that’s our**m**value. We have the**x**coordinate of the point which is a**1**we just substitute this into the original curve equation in question to find the**y**value that is; Now we know the point is**(1, 7)**. From here we can use the y – y_{1}= m(x – x_{1}) to form the tangent equation at that point. y = 6x + 1 is our final answer.### Finding the equation of a normal

This is exactly the same except instead of using the tangent gradient we use the normal gradient as described earlier. This gives us the gradient of the tangent which is a**3**. This means the normal gradient is**-1/3**. Now we continue to finding the full coordinate by substituting the**x**given value in the original equation of the curve. So our**y**coordinate is a**-2**this gives us a point of**(3, -2)**Now we use the equation as above y – y_{1}= m(x – x_{1}) The equation for the normal is**3y + x = 9**