# Tangents and Normals at a given point

**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 $-1/m$. See more on perpendicular straight lines.

Tangents and normals, like any other straight lines, have equations of the form $y=mx+c$. 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 $dy/dx$. Substituting in an $x$-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 $y$-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 $m$, the normal has a gradient of $-\frac{1}{m}$. The $y$-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 of Tangents & Normals

### Example 1

Find the equation of the tangent to

$y=x^2$

at the point $(3,9)$.

The derivative is given by $\frac{dy}{dx}=2x$ and so the gradient when $x=3$ is $m=6$. So far, we have that the equation of the tangent is $y=6x+c$. To find $c$, we use the fact that the point $(3,9)$ is on the tangent as well as the curve (it is the only point) and so when $x$ is 3, we should get $y=9$: $9=6\times 3+c$. This gives $c=-9$ and the equation of the tangent is therefore $y=6x-9$.

### Example 2

The curve $C$ has equation $y=2x^3+x^2-3x-7$. The point $P$ with $x$-coordinate -0.5 lies on $C$ and the tangent to $C$ at $P$ is parallel to the straight line $2y-3x+4=0$. Find the equation of the tangent at $P$ in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers.

The derivative is given by $\frac{dy}{dx}=6x^2+2x-3$. At $x=-0.5$, $dy/dx=-2.5$. This is the gradient of the curve and hence the gradient of the tangent at $x=-0.5$. The tangent will have equation $y=-\frac{5}{2}x+c$, where we need to find the $y$-coordinate in order to find $c$. The $y$-coordinate at $x=-0.5$ is $y=-5.5$ and so $-\frac{11}{2}=-\frac{5}{2}\times -\frac{1}{2}+c$ giving $c=-\frac{27}{4}$. It follows that the equation of the tangent is $y=-\frac{5}{2}x-\frac{27}{4}$ or rearranging into the required form: $10x+4y+27=0$.

### Example 3

Find the equation of the normal to

$y=x^2$

at the point $(3,9)$.

We saw in the above example that the gradient of the tangent is 6, and so the gradient of the normal is $m=-1/6$. So far we have that the equation of the normal is $y=-\frac{1}{6}x+c$. To find $c$, we use the fact that the point $(3,9)$ is on the normal as well as the curve (the only point that the curve, tangent and normal have in common) and so when $x$ is 3, we should get $y=9$: $9=-\frac{1}{6}\times 3+c$. This gives $c=9.5$ or $19/2$ and the equation of the tangent is therefore $y=-\frac{1}{6}x+\frac{19}{2}$.

### Example 4

The points $A$ and $B$ lie on the curve with equation $y=f(x)$. Given that $f'(x)=\frac{{x}}{3}-\frac{2}{{x}}+\frac{7}{3}$ and the normals at $A$ and $B$ are both parallel to the line $2y+x=5$, find the $x$-coordinates of $A$ and $B$.

The line $2y+x=5$ can be rewritten as $y=-\frac{1}{2}x+\frac{5}{2}$. If the normals are parallel to this line, it means that the normals have gradient $-1/2$. It follows that the tangents and the curve at points $A$ and $B$ all have gradient 2. Hence, we must solve $f'(x)=2$:

$\frac{x}{3}-\frac{2}{x}+\frac{7}{3}=2$ i.e. $\frac{1}{3}x^2-2+\frac{7}{3}x=2x$

or $x^2+x-6=(x-2)(x+3)=0$. It follows that the $x$-coordinates of $A$ and $B$ are $x=2$ and $x=-3$.