Finding equations of tangents and normals to curves at given points.
Answer at least 3 of 3 correctly to complete this section.
You have learnt to differentiate functions and find gradients. Now you will use those skills to answer one of the most classic exam questions in all of A-Level Maths: find the equation of the tangent (or normal) to a curve at a given point.
A tangent is the straight line that just touches a curve at a point, going in the same direction as the curve. A normal is the straight line perpendicular to the tangent at that point. Together, they connect differentiation to coordinate geometry — two of the biggest topics at A-Level.
This type of question appears on virtually every A-Level paper. The method is always the same three steps, so once you have it down, it becomes reliable marks.
To find the equation of the tangent to y=f(x) at the point where x=a:
y−y1=m(x−x1)
where m is the gradient and (x1,y1) is the point on the curve.
Why point-gradient form? You could rearrange to y=mx+c and find c by substitution, but the point-gradient form is quicker and less error-prone. Most mark schemes accept the answer in either form.
Find the equation of the tangent to y=x3−2x+1 at the point where x=2.
Step 1: Differentiate: dxdy=3x2−2.
At x=2: m=3(4)−2=10.
Step 2: Find the y-coordinate: y=8−4+1=5. Point: (2,5).
Step 3: Tangent equation: y−5=10(x−2)
y=10x−15
The point (1,3) lies on the curve y=x2+2x. Find the equation of the tangent at this point.
Step 1: Check: 1+2=3 ✓. The point is on the curve.
dxdy=2x+2. At x=1: m=4.
Step 2: We already have the point (1,3).
Step 3: y−3=4(x−1)⟹y=4x−1
Find the point on y=x2−4x+7 where the tangent has gradient 6.
dxdy=2x−4. Set equal to 6:
2x−4=6⟹x=5
y=25−20+7=12. The point is (5,12).
Watch out: A common mistake is substituting the x-value into dxdy to find y. You must substitute into the original equation y=f(x). The derivative gives you the gradient, not the y-coordinate.
Answer at least 3 of 4 correctly to complete this section.
The normal to a curve at a point is the straight line perpendicular to the tangent at that point.
If two lines are perpendicular, their gradients multiply to give −1:
m1×m2=−1
So if the tangent has gradient m, the normal has gradient:
mnormal=−m1
This is called the negative reciprocal.
| Tangent gradient m | Normal gradient −m1 |
|---|---|
| 2 | −21 |
| −3 | 31 |
| 41 | −4 |
| −52 | 25 |
Watch out: The negative reciprocal of 3 is −31, not −3. You must flip the fraction and change the sign. Many students only do one of these steps.
Find the equation of the normal to y=x3−x at the point (1,0).
Step 1: dxdy=3x2−1. At x=1: mtangent=2.
Step 2: mnormal=−21
Step 3: Normal equation: y−0=−21(x−1)
y=−21x+21
The curve C has equation y=4x−x2. At the point P(1,3):
(a) Find the equation of the tangent at P.
(b) Find the equation of the normal at P.
(c) The normal meets the x-axis at point Q. Find the coordinates of Q.
(a) dxdy=4−2x. At x=1: m=2.
Tangent: y−3=2(x−1)⟹y=2x+1
(b) mnormal=−21
Normal: y−3=−21(x−1)⟹y=−21x+27
(c) Set y=0: 0=−21x+27⟹x=7.
Q=(7,0).
Answer at least 3 of 4 correctly to complete this section.
The method is always the same three steps. The only thing that changes is the differentiation technique you need.
Find the equation of the tangent to y=sinx+2cosx at x=0.
Step 1: dxdy=cosx−2sinx. At x=0: m=1−0=1.
Step 2: y=sin0+2cos0=0+2=2. Point: (0,2).
Step 3: Tangent: y−2=1(x−0)⟹y=x+2
Find the equation of the normal to y=e2x at x=0.
Step 1: dxdy=2e2x. At x=0: mtangent=2e0=2.
mnormal=−21
Step 2: y=e0=1. Point: (0,1).
Step 3: Normal: y−1=−21(x−0)⟹y=−21x+1
Find the equation of the tangent to y=(3x−1)4 at x=1.
Step 1: dxdy=4(3x−1)3⋅3=12(3x−1)3 (chain rule).
At x=1: m=12(2)3=96.
Step 2: y=(3−1)4=16. Point: (1,16).
Step 3: Tangent: y−16=96(x−1)
y=96x−80
Find the point on y=x3−3x2+2 where the tangent is parallel to y=9x+1.
Parallel lines have the same gradient, so the tangent gradient must be 9.
dxdy=3x2−6x=9
3x2−6x−9=0⟹x2−2x−3=0⟹(x−3)(x+1)=0
x=3: y=27−27+2=2. Point: (3,2).
x=−1: y=−1−3+2=−2. Point: (−1,−2).
There are two points where the tangent is parallel to y=9x+1.
Exam tip: “Parallel to the line y=mx+c” means the gradient equals m. “Perpendicular to y=mx+c” means the gradient equals −m1. Read the question carefully.
These three examples follow the same method but gradually remove the scaffolding. By the third one, you are working independently.
Example A (fully worked): Find the equation of the tangent to y=ex+x2 at x=0.
Step 1: dxdy=ex+2x. At x=0: m=e0+0=1.
Step 2: y=e0+0=1. Point: (0,1).
Step 3: Tangent: y−1=1(x−0)⟹y=x+1
Example B (one step missing — you fill it in): Find the equation of the tangent to y=lnx+x3 at x=1.
Step 1: dxdy=x1+3x2. At x=1: m=1+3=4.
Step 2: Find the y-coordinate yourself: y=?
y=ln1+13=0+1=1. Point: (1,1).
Step 3: Write the tangent equation using m=4 and the point you found.
y−1=4(x−1)⟹y=4x−3
Example C (independent): Find the equation of the tangent to y=sinx+e2x at x=0.
Step 1: dxdy=cosx+2e2x. At x=0: m=1+2=3.
Step 2: y=sin0+e0=0+1=1. Point: (0,1).
Step 3: Tangent: y−1=3(x−0)⟹y=3x+1
Answer at least 3 of 4 correctly to complete this section.
The curve C has equation y=x8+2x, for x>0.
(a) Express y in the form axp+bxq where a, b, p, q are constants.
(b) Find dxdy.
(c) Find the equation of the tangent to C at the point where x=4.
(d) Find the equation of the normal to C at the point where x=4.
(a) y=8x−1/2+2x1/2
(b) dxdy=−4x−3/2+x−1/2=−x3/24+x1
(c) At x=4:
Gradient: dxdy=−84+21=−21+21=0
y-coordinate: y=28+2(2)=4+4=8. Point: (4,8).
Tangent gradient is 0, so the tangent is horizontal: y=8.
(d) Since the tangent is horizontal (m=0), the normal is vertical: x=4.
The curve C has equation y=x3−4x. The tangent at P(2,0) meets the curve again at Q. Find the coordinates of Q.
Step 1: dxdy=3x2−4. At x=2: m=12−4=8.
Tangent: y−0=8(x−2)⟹y=8x−16
Step 2: Find where the tangent meets the curve again. Set x3−4x=8x−16:
x3−12x+16=0
We know x=2 is a root (the tangent touches at P). Since the tangent is tangent to the curve at P, x=2 is a repeated root. Factor out (x−2)2:
x3−12x+16=(x−2)2(x+4)=0
So x=−4. Then y=8(−4)−16=−48.
Q=(−4,−48).
Answer at least 3 of 4 correctly to complete this section.
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