Tangents & Normals (AQA GCSE Further Maths): Revision Note

• A good sketch of the curve and the tangent at a point can help you spot if the tangent will have a positive or negative gradient; helping you to check your answer

Work out the equation of the tangent to the curve $y=2x^2-6x+10$ at the point where $x=1$.
Write your answer in the form $y=mx+c$.

Find the derivative of the curve.

$\begin{array}{rcl}\frac{dy}{dx}& =& 4x-6\end{array}$ 

To find the gradient of the curve at the point where $x=1,$ substitute $x=1$ into the derivative of the curve.

$\begin{array}{rcl}m& =& 4\left(1\right)-6=-2\end{array}$

This is the same as the gradient of the tangent to the curve at the point where $x=1$, so the equation of the line is in the form $y=-2x+c.$ 

To find the value of $c$ we will need to know the full coordinate at the point where $x=1.$ We can find this by substituting $x=1$ into the equation for the curve (be careful to substitute it into the original equation and not your differentiated version).

$y=2{\left(1\right)}^2-6\left(1\right)+10$

Simplify to find the value of $y$ and hence, the full coordinate.

$\begin{array}{rcl}y& =& 2-6+10=6\end{array}$

Substitute $x=1$ and $y=6$ into the equation of the line.

$6=-2\left(1\right)+c$

Solve this equation to find $c.$
$\begin{array}{rcl}6& =& -2+c\\ c& =& 8\end{array}$

Write out the equation of the tangent to the curve in the form given in the question.

$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{8}$

• It sounds obvious, but read the question carefully to see if you need to find a normal or a tangent! (and check again at the end!)

  • It is a very common mistake, especially under exam pressure

Work out the equation of the normal to the curve $y=\frac{4}{x}$ at the point where $x=1$.
Give your answer in the form $ax+by+c=0$ where $a$, $b$, and $c$ are integers.

Find the derivative of the curve, you will need to rewrite the equation of the curve using index form first.

$\begin{array}{rcl}y& =& 4x^{-1}\\ \frac{dy}{dx}& =& -4x^{-2}=-\frac{4}{x^2}\end{array}$ 

To find the gradient of the curve at the point where $x=1,$ substitute $x=1$ into the derivative of the curve.

$\begin{array}{rcl}g& =& -\frac{4}{1^2}=-4\end{array}$

This is the same as the gradient of the tangent to the curve at the point where $x=1$, to find the gradient of the normal to the curve at the point where $x=1$ find the negative reciprocal of $g.$

$\begin{array}{rcl}m& =& -\frac{1}{g}=-\frac{1}{-4}=\frac{1}{4}\end{array}$

So the equation of the normal is in the form $y=\frac{1}{4}x+c.$ 

To find the value of $c$ we will need to know the full coordinate at the point where $x=1.$ We can find this by substituting $x=1$ into the equation for the curve (be careful to substitute it into the original equation and not your differentiated version).

$y=\frac{4}{1}=4$

Substitute $x=1$ and $y=4$ into the equation of the normal.

$4=\frac{1}{4}\left(1\right)+c$

Solve this equation to find $c.$

$\begin{array}{rcl}4& =& \frac{1}{4}+c\\ c& =& \frac{15}{4}\end{array}$

Write out the equation of the normal to the curve in the form $\begin{array}{rcl}y& =& mx+c\end{array}$ and then rearrange to the form given in the question.

$y=\frac{1}{4}x+\frac{15}{4}$

Multiply each term by 4.

$4y=x+15$

Subtract $4y$ from both sides.

$\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{+}\mathbf{15}\mathbf{=}\mathbf{0}$