Tangents and Normals to a Curve – A Level Maths Revision

Finding the derivative of a function allows you to find the gradient of any point on a curve. If you can find the gradient, you can find the equation of tangents and normals to a curve at a given point. This A Level revision blog will explain how to do this.

You’ll need an understanding of what differentiation is and how to differentiate a function – if you’re not sure on this, check out our blogs on Differentiation From First Principles and Differentiating Functions.

If you’d like to see this blog in PDF or PowerPoint formats, click here. If you’d like to practice some of the required prior knowledge, try these multiple-choice questions. If you think you’ve mastered differentiation and want to try something harder, try these exam-style questions.


The tangent to a curve at a given point, P, is the straight line that has the same gradient as the curve at P and touches the curve at P. You can use differentiation to find the gradient of a curve, and therefore the gradient of the corresponding tangent, at a certain point.

Example Question 1

Find the equation of the tangent to the curve 𝑦 = π‘₯4 – 6π‘₯2 + 11π‘₯ + 35 at the point (-2, 5).

First, you need to differentiate the equation of the curve to find the gradient function. Remember, to differentiate π‘₯𝑛, multiply by 𝑛 then subtract 1 from 𝑛.

\begin{aligned} &y = x^4 - 6x^2+11x + 35\\&\frac{dy}{dx}=4x^3-12x + 11 \end{aligned}

This gives you a function for the gradient which varies with π‘₯. To find the gradient at a particular point, substitute in an π‘₯-coordinate – in this case, π‘₯ = -2.

\begin{aligned} \text{gradient} &=4(-2)^3-12(-2)+11 \\ &=-32 + 24 + 11 \\ &=3 \end{aligned}

Now that you know the gradient of the curve at the point (-2, 5), you can find the equation of the tangent at the point. The gradients will be the same and the straight line will go through the point (-2, 5). The quickest way to find the equation of a straight line is by using 𝑦 – 𝑦1 = π‘š(π‘₯ – π‘₯1), where (π‘₯1, 𝑦1) are the coordinates of a point the line passes through.

\begin{aligned} & y - y_1=m(x-x_1)\\&y-5=3(x--2) \\ & y-5=3x+6\\&y=3x+11 \end{aligned}

You may be able to use your calculator to check your answer, although not every calculator has this function. It can be found in the top section of your calculator, above the definite integral button.

Choosing this function will give the following:

\frac{d}{dx}(\square)_{x = {\square}} 

The equation of the curve should be entered in the first box, in brackets. Then, use the right arrow to navigate to the second box and enter the value of π‘₯. To find the gradient of the curve 𝑦 = π‘₯4 – 6π‘₯2 + 11π‘₯ + 35 at the point (-2, 5), your calculator display should read:

\frac{d}{dx}(x^4 - 6x^2+11x+35)_{x=-2} 

You should only use this to check your work – you should show full working even if the question doesn’t specifically ask for it. However, it is a quick way to confirm your answer.


You may also be asked to find the equation of the normal to a curve at a particular point. The normal is perpendicular to the tangent.

The gradient of the normal is the negative reciprocal of the gradient of the tangent (or curve) at the point. In other words, the gradient of the tangent, π‘šπ‘‘, and the gradient of then normal, π‘šπ‘›, will have the property:

m_t \times m_n = -1

Example Question 2

Find the equation of the normal to the curve 𝑦 = 6π‘₯3 – 2π‘₯2 – 5π‘₯ + 2 at the point \left( \frac{1}{2}, -\frac{1}{4} \right) . Give your answer in the form π‘Žπ‘₯ + 𝑏𝑦 + 𝑐 = 0 where π‘Ž, 𝑏 and 𝑐 are integers.

Just like in example 1, start by differentiating the equation of the curve:

\begin{aligned} &y=6x^3-2x^2-5x+2\\&\frac{dy}{dx}=18x^2-4x-5 \end{aligned}

Then, find the gradient of the curve when π‘₯ = \frac{1}{2} . This will be the gradient of the tangent, π‘šπ‘‘. To find the gradient of the normal, π‘šπ‘›, you need to find the negative reciprocal of π‘šπ‘‘:

\begin{aligned} m_t &= 18\left( \frac{1}{2} \right)^2 - 14 \left(\frac{1}{2} \right) - 5 \\ &=\frac{9}{2}-2-5\\&=-\frac{5}{2} \end{aligned}

Now that you have the gradient of the tangent, you can find the gradient of the normal:

\begin{aligned} &m_t \times m_n = -1 \\ &-\frac{5}{2} \times m_n = -1 \\ &m_n=\frac{2}{5} \end{aligned}

The gradient of the normal is \frac{2}{5} and it passes through the point \left( \frac{1}{2}, -\frac{1}{4} \right) . This is all the information you need to find the equation of the normal:

\begin{aligned} &y - y_1 = m(x - x_1) \\ &y - -\frac{1}{4} = \frac{2}{5}(x- \frac{1}{2})  \end{aligned}

The question asks for the equation in the form π‘Žπ‘₯ + 𝑏𝑦 + c = 0 where π‘Ž, 𝑏 and 𝑐 are integers, so rearrange the equation to get it in this form:

\begin{aligned} &y +\frac{1}{4} = \frac{2}{5}(x- \frac{1}{2}) \\ &5(y+\frac{1}{4})=2(x-\frac{1}{2})\\&5y+\frac{5}{4}=2x-1\\&20y+5=8x-4\\&8x-20y-9=0 \end{aligned}


Example Question 3

The graph 𝑦 = f(π‘₯) has a tangent at the point where π‘₯ = 2 and a normal at the point where π‘₯ = \frac{1}{2} . Given that f(π‘₯) = \frac{1}{x} + π‘₯2, find the equations of the tangent and normal.

This question is slightly different to the others – in examples 1 and 2, all the terms in the curves’ equations were in the form π‘Žπ‘₯𝑛, where 𝑛 is a positive integer. In this case, some values of 𝑛 will be negative. The same method will apply but you need to make sure the terms are in the form π‘Žπ‘₯𝑛 before you differentiate. You may also have questions in which 𝑛 is a fraction.

\begin{aligned} f(x)&=\frac{1}{x} + x^2 \\ &=x^{-1}+x^2 \end{aligned}

Now that you have the function in this form, you can differentiate with respect to π‘₯. Take care with fractional and negative values.

\begin{aligned} f'(x) &= -x^{-2}+2x \\&=-\frac{1}{x^2}+2x \end{aligned}

Now, find the equation of the tangent. The question doesn’t give both coordinates so substitute π‘₯ = 2 to find the 𝑦-coordinate at that point:

\begin{aligned} y &= \frac{1}{x} +x^2 \\[0.5em] &=\frac{1}{2} + 2^2 \\[0.5em] &= \frac{9}{2} \end{aligned}

Now, find the gradient of the tangent:

\begin{aligned} f'(2) &= -\frac{1}{2^2}+2\times 2 \\[0.5em] &=-\frac{1}{4}+4 \\[0.5em] &= \frac{15}{4} \end{aligned}

Now that you have the point at which the tangent meets the curve and the gradient, you can use 𝑦 – 𝑦1 = π‘š(π‘₯ – π‘₯1) to find the equation of the tangent:

\begin{aligned} & y - y_1 = m(x-x_1) \\[0.5em] & y - \frac{9}{2}=\frac{15}{4}(x - 2) \\[0.5em] & 4(y-\frac{9}{2})=15(x-2) \\[0.5em] &4y-18=15x-30 \\[0.5em]&15x-4y-12=0\end{aligned}

Now, repeat this process to find the equation of the normal. First, find the 𝑦-coordinate when π‘₯ = \frac{1}{2}:

\begin{aligned} y &= \frac{1}{\frac{1}{2}} + (\frac{1}{2})^2 \\[0.5em]&=2+\frac{1}{4} \\[0.5em] &=\frac{9}{4} \end{aligned}

Then, find the gradient of the tangent at the point π‘₯ = \frac{1}{2}:

\begin{aligned} f'\left(\frac{1}{2}\right) &= -\frac{1}{\left(\frac{1}{2}\right)^2}+2\times\frac{1}{2}\\&=-4+1\\&=3\end{aligned}

This is the gradient of the tangent, so the normal will have gradient \frac{1}{3}.

Finally, use 𝑦 – 𝑦1 = π‘š(π‘₯ – π‘₯1) to find the equation of the normal:

\begin{aligned} &y-\frac{9}{4}=\frac{1}{3}(x-\frac{1}{2})\\[0.25em]&3y-\frac{27}{4}=x-\frac{1}{2}\\[0.25em]&12y-27=4x-2\\[0.25em]&4x-12y+25=0\end{aligned}


Practice Questions

1. Find the gradient of each of the graphs below at the point π‘₯ = 3.

a. y = 3x^2 + 2x - 5

b. y = \frac{4}{x^2}

c. y = (3x-2)(1-x)

d. y = \sqrt{x^3}+2x^4

2. Find the equation of the tangent to the curve 𝑦 = π‘₯3 + 3.5π‘₯2 – 2π‘₯ + 5 at the point (-2, 15). Give your answer in the form 𝑦 = π‘šπ‘₯ + 𝑐.

3. Find the equation of the normal to the curve 𝑦 = \frac{1}{x^2} -7π‘₯ at the point (-1, 8). Give your answer in the form 𝑦 = π‘šπ‘₯ + 𝑐.

4. A tangent to the curve 𝑦 = π‘₯3 + 2π‘₯2 – 3 with the equation π‘₯ + 𝑦 + 3 = 0 meets the curve at two points.
a. Find the coordinates of the two points.
b. Determine the point at which π‘₯ + 𝑦 + 3 = 0 is a tangent to the curve.

5. The tangent to the curve y = \frac{1}{x^3}+\frac{1}{x\sqrt{x}}-\frac{x}{2} at the point (1,\frac{3}{2}) meets the π‘₯- and 𝑦-axis at the points P and Q respectively. Find the area of the triangle OPQ, where O is the origin.


Answers

1. Find the gradient of each of the graphs below at the point π‘₯ = 3.

a. y = 3x^2 + 2x - 5

\boldsymbol{\frac{dy}{dx}=6x+2}
\boldsymbol{\begin{aligned}\textbf{gradient}&=6\times 3 + 2 \\ &=20 \end{aligned}}

b. y = \frac{4}{x^2}

\boldsymbol{\frac{dy}{dx}=-\frac{8}{x^3}}
\boldsymbol{\begin{aligned}\textbf{gradient}&=-\frac{8}{3^3}\\& = -\frac{8}{27} \end{aligned}}

c. y = (3x-2)(1-x)

\boldsymbol{\begin{aligned}&y=-3x^2+5x-2\\&\frac{dy}{dx}=-6x+5\end{aligned}}
\boldsymbol{\begin{aligned}\textbf{gradient}&=-6\times 3 + 5 &\\&= -13 \end{aligned}}

d. y = \sqrt{x^3}+2x^4

\boldsymbol{y=x^{\frac{3}{2}}+2x^4}
\boldsymbol{\begin{aligned}\frac{dy}{dx}&=\frac{3}{2}x^{\frac{1}{2}}+8x^3\\&=\frac{3}{2}\sqrt{x}+8x^3 \end{aligned}}
\boldsymbol{\begin{aligned}\textbf{gradient}&=\frac{3}{2}\sqrt{3}+8\times 3^3\\ &= \frac{3\sqrt{3}}{2}+216 \end{aligned}}

2. Find the equation of the tangent to the curve 𝑦 = π‘₯3 + 3.5π‘₯2 – 2π‘₯ + 5 at the point (-2, 15). Give your answer in the form 𝑦 = π‘šπ‘₯ + 𝑐.

\boldsymbol{y=x^3+3.5x^2-2x+5}
\boldsymbol{\frac{dy}{dx}=3x^2+7x-2}

\boldsymbol{\begin{aligned}m_t&=3x^2+7x-2\\&=3(-2)^2+7(-2)-2\\&=12-14-2\\&=-4\end{aligned}}

\boldsymbol{\begin{aligned}&y-y_1=m(x-x_1)\\&y-15=-4(x--2)\\&y-15=-4x-8\\&y=-4x+7 \end{aligned}}

3. Find the equation of the normal to the curve 𝑦 = \frac{1}{x^2} -7π‘₯ at the point (-1, 8). Give your answer in the form 𝑦 = π‘šπ‘₯ + 𝑐.

\boldsymbol{\begin{aligned}y&=\frac{1}{x^2}-7x\\&=x^{-2}-7x\end{aligned}}
\boldsymbol{\frac{dy}{dx}=-2x^{-3}-7}

\boldsymbol{\begin{aligned}m_t&=-2x^{-3}-7\\&=-2(-1)^{-3}-7\\&=-5\\m_n&=\frac{1}{5}\end{aligned}}

\boldsymbol{\begin{aligned}&y-y_1=m(x-x_1)\\&y-8=\frac{1}{5}(x--1)\\&y-8=\frac{1}{5}x+\frac{1}{5}\\&y=\frac{1}{5}x+\frac{41}{5}\end{aligned}}

4. A tangent to the curve 𝑦 = π‘₯3 + 2π‘₯2 – 3 with the equation π‘₯ + 𝑦 + 3 = 0 meets the curve at two points.
a. Find the coordinates of the two points.

\boldsymbol{\begin{aligned}&x + y + 3 = 0 \\ &y = -x-3\end{aligned}}

\boldsymbol{\begin{aligned}&\textbf{Solving simultaneously:}\\&-x-3=x^3+2x^2-3\\&x^3+2x^2+x=0\\&x(x^2+2x+1)=0\\&x(x+1)(x+1)=0\\&x=0\textbf{ and } y=-3\\&x=-1 \textbf{ and } y=-2 \end{aligned}}

The curve and the line meet at (0,-3) and (-1,-2).

b. Determine the point at which π‘₯ + 𝑦 + 3 = 0 is a tangent to the curve.

Since 𝑦 = -π‘₯ – 3, the gradient of the tangent is -1. To find the point at which the line is a tangent to the curve, find the point at which the gradient of the curve is -1.

\boldsymbol{\begin{aligned}&y=x^3+2x^2-3 \\ &\frac{dy}{dx}=3x^2+4x \\ &3x^2 + 4x = -1 \\ &3x^2+4x+1 =0 \\&(3x+1)(x+1)=0\\&x=-\frac{1}{3}\textbf{ or } x=-1 \end{aligned}}

Only one of these is also a point where the curve and straight line meet: π‘₯ = -1.

Therefore, the line π‘₯ + 𝑦 + 3 = 0 is a tangent to the curve 𝑦 = π‘₯3 + 2π‘₯2 – 3 at the point (-1, -2).

5. The tangent to the curve y = \frac{1}{x^3}+\frac{1}{x\sqrt{x}}-\frac{x}{2} at the point (1,\frac{3}{2}) meets the π‘₯- and 𝑦-axis at the points P and Q respectively. Find the area of the triangle OPQ, where O is the origin.

\boldsymbol{\begin{aligned}y&=\frac{1}{x^3}+\frac{1}{x\sqrt{x}}-\frac{x}{2}\\[0.5em]&=\frac{1}{x^3}+\frac{1}{\sqrt{x^3}}-\frac{x}{2}\\[0.5em]&=x^{-3}+x^{-\frac{3}{2}}-\frac{1}{2}x\end{aligned}}

\boldsymbol{\begin{aligned}\frac{dy}{dx}=3x^{-4}-\frac{3}{2}x^{-\frac{5}{2}}-\frac{1}{2}\end{aligned}}

\boldsymbol{\begin{aligned}m_t&=-3(1)^{-4}-\frac{3}{2}(1)^{-\frac{5}{2}}-\frac{1}{2}\\&=-5\end{aligned}}

\boldsymbol{\begin{aligned}y-y_1=m(x-x_1)\\&y-\frac{3}{2}=-5(x-1)\\&y-\frac{3}{2}=-5x+5\\&y=-5x+\frac{13}{2}\end{aligned}}

When π‘₯ = 0, 𝑦 = \boldsymbol{\frac{13}{2}} and when 𝑦 = 0, π‘₯ = \boldsymbol{ \frac{13}{10}} .

Therefore, the triangle OPQ has height \boldsymbol{\frac{13}{2}} and base \boldsymbol{\frac{13}{10}}.

\boldsymbol{\begin{aligned}\textbf{Area of Triangle}&=\frac{1}{2} \times \textbf{base} \times \textbf{height}\\[0.5em]&=\frac{1}{2}\times\frac{13}{10}\times\frac{13}{2} \\[0.5em]&= \frac{169}{40} \textbf{ square units}\end{aligned}}

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