5.4 Tangent Line And Normal Line

Remember: A straight line with

  • gradient $m$
  • passing through point $(x_0, y_0)$

has the equation:

$y - y_0 = m(x - x_0)$

For example, the line passing through $A(1, 2)$ with gradient $m=3$ has the equation:

$y - 2 = 3(x - 1)$

Consider a function $y = f(x)$ and some point $x_0$. Then we also know $y_0 = f(x_0)$.

x₀ y₀
x₀ y₀ tangent line normal line

We know that the gradient of the graph at $(x_0, y_0)$ is $m_T = f'(x_0)$.

We define:

  • TANGENT LINE at $x_0$:
    the line with gradient $m_T$ which passes through $(x_0, y_0)$.
  • NORMAL LINE at $x_0$:
    the perpendicular line to the tangent at $(x_0, y_0)$. Its gradient is:
    $m_N = -\dfrac{1}{m_T}$

The point $(x_0, y_0)$ is also known as the point of contact.

METHODOLOGY

Given $y = f(x)$ and some point $x = x_0$:

  • We find the point of contact $(x_0, y_0)$ since $y_0 = f(x_0)$.
  • We find $f'(x)$.
  • We find $m_T = f'(x_0)$ and so $m_N = -\dfrac{1}{m_T}$.

The equations of the two lines are:

TANGENT LINE:
$y - y_0 = m_T(x - x_0)$
NORMAL LINE:
$y - y_0 = m_N(x - x_0)$

EXAMPLE 1

Consider the function $f(x) = x^2$.
Find the equations of the tangent line and the normal line at $x=3$.

Solution:
The point of contact is $(3, 9)$ (since $f(3) = 9$).
The gradient function is $f'(x) = 2x$. Thus,
$m_T = f'(3) = 6$ and $m_N = -\dfrac{1}{6}$.

The tangent line is: $y - 9 = 6(x - 3)$
The normal line is: $y - 9 = -\dfrac{1}{6}(x - 3)$

If they ask us to express them in the form $y = mx + c$, we obtain:
Tangent line:
$y - 9 = 6(x - 3) \iff y - 9 = 6x - 18 \iff \mathbf{y = 6x - 9}$
Normal line:
$y - 9 = -\dfrac{1}{6}(x - 3) \iff y - 9 = -\dfrac{1}{6}x + \dfrac{1}{2} \iff \mathbf{y = -\dfrac{1}{6}x + \dfrac{19}{2}}$

NOTICE (Alternative Way)

An alternative way to obtain the tangent and the normal lines is to use the formulas:

TANGENT LINE: $y = m_T x + c$
NORMAL LINE: $y = m_N x + c$

The point $(x_0, y_0)$ helps us to find the constant $c$.

In the previous example, since $m_T=6$ and the point is $(3, 9)$:
Tangent line: $y = 6x + c$
At $(3, 9)$: $18 + c = 9 \iff c = -9$, thus $\mathbf{y = 6x - 9}$

Normal line: $y = -\dfrac{1}{6}x + c$
At $(3, 9)$: $-\dfrac{1}{6}(3) + c = 9 \iff c = 9 + \dfrac{1}{2} \iff c = \dfrac{19}{2}$, thus $\mathbf{y = -\dfrac{1}{6}x + \dfrac{19}{2}}$

EXAMPLE 2

Consider the function $f(x) = 5x^3 - 2x + 1$.
Find the tangent lines to the curve which are parallel to the line $L: y = 13x + 8$.

Solution:
A tangent line parallel to $L$ must have gradient $m_T = 13$.
But $f'(x) = 15x^2 - 2$, so:
$15x^2 - 2 = 13 \iff 15x^2 = 15 \iff x^2 = 1 \iff x = 1$ or $x = -1$

Hence, we will have two parallel lines, at the points $x = 1$ and $x = -1$ with gradient $m = 13$. The points of contact are $(1, 4)$ and $(-1, -2)$.
  • At $(1, 4)$: $y - 4 = 13(x - 1) \iff y - 4 = 13x - 13 \iff \mathbf{y = 13x - 9}$
  • At $(-1, -2)$: $y + 2 = 13(x + 1) \iff y + 2 = 13x + 13 \iff \mathbf{y = 13x + 11}$

NOTICE (Zero Gradient)

Suppose that the gradient at $(x_0, y_0)$ is $m_T = 0$. Then:

  • The tangent line is a horizontal line with equation $\mathbf{y = y_0}$.
  • The normal line is a vertical line with equation $\mathbf{x = x_0}$.

EXAMPLE 3

Consider the function $f(x) = x^2 - 4x + 5$.
Find the equations of the tangent line and the normal line at $x=2$.

x y x=2 y=1 2 1
Solution:
It is $f'(x) = 2x - 4$.
At $x = 2$, $y = 1$, thus the point of contact is $(2, 1)$.
$m_T = 0$ ($m_N$ is not defined).

Tangent line: the horizontal line $\mathbf{y = 1}$
Normal line: the vertical line $\mathbf{x = 2}$ (look at the graph!)

Observation for Trickier Questions

At the point of contact between $f(x)$ and a tangent line $y = mx + c$:

  • functions are equal: $f(x) = y$
  • derivatives are equal: $f'(x) = m$

Note: A vertical line has no gradient. If it passes through $(x_0, y_0)$ it has equation $x = x_0$.

EXAMPLE 4

The line $y = mx - 3$ is tangent to the curve $f(x) = x^4 - x$. Find $m$.

Solution:
At the point of contact $x$:
Functions are equal: $x^4 - x = mx - 3$
Derivatives are equal: $4x^3 - 1 = m$

Hence, replacing $m$:
$x^4 - x = (4x^3 - 1)x - 3 \iff x^4 - x = 4x^4 - x - 3 \iff 3x^4 = 3 \iff x^4 = 1 \iff x = \pm 1$

If $x = 1$, then $\mathbf{m = 3}$.
If $x = -1$, then $\mathbf{m = -5}$.

EXAMPLE 5

Consider the function $f(x) = x^4 - x$. Find the tangent lines passing through the point $(0, -3)$.
[Notice that the point is not on the curve itself.]

Method A:
A line passing through the point $(0, -3)$ has the form:
$y + 3 = m(x - 0) \implies y = mx - 3$
This is in fact Example 4 above. We found two solutions:
If $x = 1$, then $m = 3$ and the tangent line is $\mathbf{y = 3x - 3}$.
If $x = -1$, then $m = -5$ and the tangent line is $\mathbf{y = -5x - 3}$.
Method B: (First find the tangent line at an arbitrary point of contact $x=a$)
At any point $(a, f(a))$, i.e., $(a, a^4 - a)$:
$f'(x) = 4x^3 - 1 \implies m_T = 4a^3 - 1$.
Therefore,
$y - (a^4 - a) = (4a^3 - 1)(x - a) \implies y = (4a^3 - 1)x - 3a^4$.

Because this line must strictly pass through $(0, -3)$, we substitute $x=0, y=-3$:
$-3 = (4a^3 - 1)(0) - 3a^4 \implies -3 = -3a^4 \iff a^4 = 1 \iff a = \pm 1$

For $a = 1$, we systematically obtain the equation $\mathbf{y = 3x - 3}$.
For $a = -1$, we systematically obtain the equation $\mathbf{y = -5x - 3}$.