Tangents and normals
Tangents and normals are lines at a given point on a curve. A tangent runs parallel with the curve at the point whereas the normal is perpendicular to the curve. It follows that if the curve has gradient m, the tangent also has gradient m and the normal has gradient 
. See more on perpendicular straight lines.
Tangents and normals, like any other straight lines, have equations of the form 
. Their equations can be found just like any other straight lines. However, you are likely to need the derivative by differentiating when finding the gradient of tangents and normals to a curve.
Tangents
As mentioned above, the TANGENT to a curve at a given point has the same gradient as the curve at that point. In order to find the gradient of the curve, we require the derivative. This is found by differentiating and obtaining an expression for 
. Substituting in an 
-value will give the gradient of the curve and hence the gradient of the tangent. To find the full equation of the tangent, it remains to find the 
-intercept of the tangent. This can be achieved using the coordinates of the point that the tangent and the curve have in common.
See Examples 1 and 2. Page 13 onwards of the StudyWell Differentiation eGuide has more on Tangents & Normals including some exam-style questions.
Normals
The NORMAL to a point on a curve is the line that is perpendicular to the tangent at the given point. If the tangent has a gradient of 
, the normal has a gradient of . The -intercept of the normal, which is different from that of the tangent, can also be found using the coordinates of the given point.
See Examples 3 and 4. Page 13 onwards of the StudyWell Differentiation eGuide has more on Tangents & Normals including some exam-style questions.
Examples
Find the equation of the tangent to
at the point 
.
Solution:
The derivative is given by 
and so the gradient when 
is 
. So far, we have that the equation of the tangent is 
. To find 
, we use the fact that the point is on the tangent as well as the curve (it is the only point) and so when is 3, we should get 
: 
. This gives 
and the equation of the tangent is therefore 
.
The curve 
has equation 
. The point 
with -coordinate -0.5 lies on and the tangent to at is parallel to the straight line 
. Find the equation of the tangent at in the form 
where 
, 
and are integers.
Solution:
The derivative is given by 
. At 
, 
. This is the gradient of the curve and hence the gradient of the tangent at . The tangent will have equation 
, where we need to find the -coordinate in order to find . The -coordinate at is 
and so 
giving 
. It follows that the equation of the tangent is 
or rearranging into the required form: 
.
Find the equation of the normal to
at the point .
Solution:
We saw in the above example that the gradient of the tangent is 6, and so the gradient of the normal is 
. So far we have that the equation of the normal is 
. To find , we use the fact that the point is on the normal as well as the curve (the only point that the curve, tangent and normal have in common) and so when is 3, we should get : 
. This gives 
or 
and the equation of the tangent is therefore 
.
The points 
and 
lie on the curve with equation 
. Given that 
and the normals at and are both parallel to the line 
, find the -coordinates of and .
Solution:
The line can be rewritten as 
. If the normals are parallel to this line, it means that the normals have gradient 
. It follows that the tangents and the curve at points and all have gradient 2. Hence, we must solve 
:
i.e. 
or 
. It follows that the -coordinates of and are 
and 
.
Videos
Finding the derivative of a polynomial function and using it to find the equation of a normal at a given point on the function’s curve.
Using the definition of parallel lines in gradients to find unknown an polynomial coefficient and subsequently the equation of a tangent.
