Tangents and Normals

The derivative of a function at a point gives the slope of the tangent line to the curve at that point. This geometric interpretation is one of the most useful and immediate applications of differentiation.

Concept	Quick Reference
Tangent line at $(x_{0}, f (x_{0}))$	y - f(x_0) = f'(x_0)(x - x_0)
Normal line at $(x_{0}, f (x_{0}))$	y - f(x_0) = -\dfrac{1}{f'(x_0)}(x - x_0)
Perpendicular slopes	$m_{1} m_{2} = - 1$
Angle between curves	$\tan α = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$

Tangents

Previously, we interpreted the derivative of a function at a point as the slope of the tangent line to the curve at that point.

The tangent line to the graph of $f$ at $(x_{0}, f (x_{0}))$ is the line through $(x_{0}, f (x_{0}))$ with slope f'(x_0).

Recall that the equation of a line passing through $(x_{0}, y_{0})$ with slope $m$ is

y - y_{0} = m (x - x_{0}) .

The equation of the tangent line to the curve at $(x_{0}, f (x_{0}))$ is therefore

y - f(x_0) = f'(x_0)(x - x_0). \tag{i}

Normal to a Curve

The line $P N$ perpendicular to the tangent line $P T$ at its point of contact with the curve is called the normal to the curve at that point.

Diagram showing a curve with tangent line PT and normal line PN at a point of contact

Suppose that $f$ is differentiable at $x_{0}$ . The normal to the function $f$ at $(x_{0}, f (x_{0}))$ is the line that passes through $(x_{0}, f (x_{0}))$ and is perpendicular to the tangent there.

Recall from analytic geometry that two lines with slopes $m_{1}$ and $m_{2}$ are perpendicular if and only if $m_{1} m_{2} = - 1$ ; a horizontal line (slope 0) is perpendicular to vertical lines (no slope).
Therefore, the normal to $f$ at $(x_{0}, f (x_{0}))$ is a line through $(x_{0}, f (x_{0}))$ with slope -1/f'(x_0). The equation of the normal is y - f(x_0) = -\frac{1}{f'(x_0)}(x - x_0).

Find the equations of the tangent and normal to the curve $y = 3 \cot x - \cot^{3} x + 1$ at $x = π / 6$ .

Solution

Let

f (x) = 3 \cot x - \cot^{3} x + 1

. Then \begin{aligned} f\left(\frac{\pi}{6}\right) &= 3\cot\left(\frac{\pi}{6}\right) - \left[\cot\left(\frac{\pi}{6}\right)\right]^3 + 1 \\ &= 3\sqrt{3} - (\sqrt{3})^3 + 1 \\ &= 1. \end{aligned} [Note that

\cot (π / 6) = \frac{\cos (π / 6)}{\sin (π / 6)} = \frac{\sqrt{3} / 2}{1 / 2} = \sqrt{3}

.] f'(x) = -3(1 + \cot^2 x) - 3\cot^2 x\left[-(1 + \cot^2 x)\right] + 0 \tag{Recall $(u^r)' = ru^{r-1}u'$} letting

u = \cot x

. [Recall that (\cot x)' = -(1 + \cot^2 x).] At

x = π / 6

the slope of the tangent line is \begin{aligned} f'\left(\frac{\pi}{6}\right) &= -3\left[1 + \cot^2\left(\frac{\pi}{6}\right)\right] + 3\cot^2\left(\frac{\pi}{6}\right)\left[1 + \cot^2\left(\frac{\pi}{6}\right)\right] \\ &= -3[1 + (\sqrt{3})^2] + 3(\sqrt{3})^2[1 + (\sqrt{3})^2] \\ &= 24. \end{aligned} The equation of the tangent line is therefore y - f\left(\frac{\pi}{6}\right) = f'\left(\frac{\pi}{6}\right)\left(x - \frac{\pi}{6}\right)

y - 1 = 24 (x - \frac{π}{6})

y = 24 x - 4 π + 1.

Since the slope of the normal at

x = π / 6

is -\dfrac{1}{f'(\pi/6)} = -\dfrac{1}{24}, the equation of the normal at the point

(π / 6, 1)

y - 1 = - \frac{1}{24} (x - \frac{π}{6}) .

The following figure shows the graph of

f

and its tangent and normal at the point

(π / 6, 1)

Graph of y = 3 cot x - cot^3 x + 1 with its tangent and normal at the point (pi/6, 1)

Find the tangent and normal to the ellipse

4 x^{2} + 9 y^{2} = 40

at the point $(- 1, 2)$ .

Solution

First we need to find the slope of the tangent line. We use implicit differentiation: 8x + 18y y' = 0 \implies y' = -\frac{4x}{9y}. Hence, m = y'\bigg|_{(-1,2)} = \left.-\frac{4x}{9y}\right|_{(-1,2)} = -\frac{4(-1)}{9(2)} = \frac{2}{9}. Therefore, the equation of the tangent line is

y - 2 = \frac{2}{9} (x + 1)

and the equation of the normal is

y - 2 = - \frac{9}{2} (x + 1) .

Ellipse 4x^2 + 9y^2 = 40 with tangent and normal at (-1, 2)

Tangent Line Through a Point Not on the Curve

Suppose that we want to determine the equation of a tangent line to a curve $y = f (x)$ that passes through a point $M (a, b)$ (see the following figure). Because $M$ is not the point of tangency, we cannot use Equation (i).

A curve with a point M not on the curve, and a tangent line from M touching the curve at point P

To solve such problems, we assume that it is possible to draw such a tangent and let $P (x_{0}, f (x_{0}))$ be the point on the given curve whose tangent line passes through $M$ . Then we try to find $P$ .

There are two methods for finding $P$ .

Method (a). The equation of the tangent line at $P$ is

y - f(x_0) = f'(x_0)(x - x_0).

Because this line passes through $M$ , the coordinates of $M$ must satisfy the equation of the tangent line; that is,

b - f(x_0) = f'(x_0)(a - x_0).

Because $a$ and $b$ are given, the only unknown in this equation is $x_{0}$ . Solving for $x_{0}$ gives the coordinates of the tangency point $P (x_{0}, f (x_{0}))$ and hence the equation of the tangent line that passes through $M$ .

Curve with tangent from external point, showing the tangency point P

Method (b). We notice that the slope of the tangent line at $P$ equals the slope of the line passing through $P$ and $M$ . In other words, we solve

\frac{f(x_0) - b}{x_0 - a} = f'(x_0)

for $x_{0}$ .

Solving the equation for $x_{0}$ might produce only one solution, more than one solution, or no solution at all.

Determine the equation(s) of the line(s) through the point $M (4, 0)$ that are tangent to the curve $y = x^{2} - 3 x$ .

Solution

**Method (a).** Let

f (x) = x^{2} - 3 x

. Since f'(x) = 2x - 3, the equation of the tangent line at a point whose

x

-coordinate is

x_{0}

becomes y - \underbrace{(x_0^2 - 3x_0)}_{f(x_0)} = \underbrace{(2x_0 - 3)}_{f'(x_0)}(x - x_0). Because this line passes through

M (4, 0)

, we have

0 - (x_{0}^{2} - 3 x_{0}) = (2 x_{0} - 3) (4 - x_{0}),

or after rearranging terms

x_{0}^{2} - 8 x_{0} + 12 = 0.

The solutions of this equation are

x_{0} = \frac{8 \pm \sqrt{8^{2} - 4 (1) (12)}}{2 (1)} = 4 \pm 2.

Therefore, there are two points of tangency:

(6, f (6)) = (6, 18) and (2, f (2)) = (2, - 2) .

The equations of the corresponding tangent lines are y - 18 = \underbrace{9}_{f'(6)}(x - 6), \quad \text{and} \quad y + 2 = \underbrace{1}_{f'(2)}(x - 2). The graphs of

f

and these two tangent lines are shown in the following figure.

Two tangent lines to the curve y = x^2 - 3x passing through M(4,0) — Two tangent lines to the curve $y = x^{2} - 3 x$ pass through $M (4, 0)$ .

**Method (b).** The slope of the tangent line at

P (x_{0}, x_{0}^{2} - 3 x_{0})

2 x_{0} - 3

. The slope of the line passing through

P (x_{0}, x_{0}^{2} - 3 x_{0})

and

M (4, 0)

\frac{x_{0}^{2} - 3 x_{0} - 0}{x_{0} - 4} .

By equating these two slopes, we get

2 x_{0} - 3 = \frac{x_{0}^{2} - 3 x_{0}}{x_{0} - 4}

x_{0}^{2} - 8 x_{0} + 12 = 0

Therefore

x_{0} = 6 or x_{0} = 2

and the equations of the tangent lines are

y - 18 = 9 (x - 6), and y + 2 = (x - 2) .

Find the equations of the tangent line to the curve $y^{2} - 6 y - 8 x - 31 = 0$ which passes through the point $(- 4, - 1)$ .

Solution

First, we notice that

(- 4, - 1)

is not on this curve as its coordinates do not satisfy the given equation. Let

(x_{0}, y_{0})

be the point whose tangent passes through

(- 4, - 1)

. The slope of the tangent line is found by implicit differentiation: 2y y' - 6y' - 8 = 0 \implies y' = \frac{4}{y - 3}. Therefore, the slope of the tangent line at

(x_{0}, y_{0})

m = \frac{4}{y_{0} - 3} .

The slope of the line passing through

(x_{0}, y_{0})

and

(- 4, - 1)

m = \frac{y_{0} + 1}{x_{0} + 4} .

These two lines are the same, so

\frac{4}{y_{0} - 3} = \frac{y_{0} + 1}{x_{0} + 4} .

Since

(x_{0}, y_{0})

lies on the curve, we can express

x_{0}

in terms of

y_{0}

y_{0}^{2} - 6 y_{0} - 8 x_{0} - 31 = 0 ⟹ x_{0} = \frac{1}{8} (y_{0}^{2} - 6 y_{0} - 31) .

Substituting into the slope equation and simplifying gives

y_{0}^{2} + 2 y_{0} - 7 = 0.

The solutions are

y_{0} = - 1 \pm 2 \sqrt{2}

. Therefore, there are two points on the curve whose tangents pass through

(- 4, - 1)

; their coordinates are

(- 2 + 2 \sqrt{2}, - 1 - 2 \sqrt{2}) and (- 2 - 2 \sqrt{2}, - 1 + 2 \sqrt{2}) .

The equations of the corresponding tangents are

y + 1 + 2 \sqrt{2} = \frac{4}{- 4 - 2 \sqrt{2}} (x + 2 - 2 \sqrt{2})

and

y + 1 - 2 \sqrt{2} = \frac{4}{- 4 + 2 \sqrt{2}} (x + 2 + 2 \sqrt{2}) .

Angle Between Two Curves

The angle between two curves at a point of intersection is the angle between their tangents at that point.

Two curves intersecting, with angle alpha between their tangent lines at the point of intersection — $α$ is the angle between the tangents of these two curves at the point of intersection.

How to find the intersection of two curves

If two curves intersect at a point

P

, then the coordinates of

P

satisfy the equations of these curves. Therefore, to find the intersections of two curves, we simultaneously solve their equations.

How to find the angle between two curves

If the slopes of their tangent lines at a point of intersection are

m_{1}

and

m_{2}

, and the angle from the curve with slope

m_{1}

to the curve with slope

m_{2}

α

, then

\tan α = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} .

Let

α = θ_{2} - θ_{1}

be the angle from the line of slope

m_{1} = \tan θ_{1}

to that of slope

m_{2} = \tan θ_{2}

. Then \begin{aligned} \tan\alpha = \tan(\theta_2 - \theta_1) &= \frac{\tan\theta_2 - \tan\theta_1}{1 + \tan\theta_1\,\tan\theta_2} \\ &= \frac{m_2 - m_1}{1 + m_1 m_2}. \end{aligned}

Two lines with angles theta_1 and theta_2 to the x-axis, with the angle alpha between them

Find the angle between the line $y = x$ and the curve $y = x^{2}$ at each intersection.

Solution

Solving simultaneously \left\{\begin{aligned} y &= x^2 \\ y &= x \end{aligned}\right. \implies x^2 = x \implies x = 0 \quad \text{or} \quad x = 1 we find the line and curve intersect at $(1, 1)$ and $(0, 0)$. The slope of

y = x

m_{1} = 1

. The slope of the curve

y = x^{2}

at any point

x

\frac{d y}{d x} = 2 x

. At $(1, 1)$ the slope of the curve is $2$, and the angle

α

from

y = x

y = x^{2}

\tan α_{1} = \frac{2 - 1}{1 + 2 \times 1} = \frac{1}{3} .

Thus

α_{1} = \arctan \frac{1}{3} = 0.3218 rad = 18.43 ° .

At $(0, 0)$, the slope of the curve is $0$, and

\tan α_{2} = \frac{0 - 1}{1 + 0} = - 1.

Therefore,

α_{2} = \arctan (- 1) = - \frac{π}{4} rad = - 45 ° .

The negative sign in the last case signifies that the acute angle is measured clockwise from the line to the curve.

Line y = x and parabola y = x^2 intersecting at (0,0) and (1,1) with angles labeled

Find the angle of intersection of the circle $x^{2} + y^{2} = 8$ and the parabola $x^{2} = 2 y$ .

Solution

Solving the equations simultaneously: \left\{\begin{aligned} &x^2 + y^2 = 8 \\ &x^2 = 2y \end{aligned}\right. \implies 2y + y^2 = 8 or

y^{2} + 2 y - 8 = 0

, giving

y = 2

y = - 4

. Since

x^{2} = 2 y \geq 0

, we need

y = 2

, giving

x = \pm 2

. The two points of intersection are $(2, 2)$ and

(- 2, 2)

. At $(2, 2)$: From the circle

2 x + 2 y \frac{d y}{d x} = 0

, so

\frac{d y}{d x} = - \frac{x}{y}

; hence the slope of the tangent to the circle is

m_{1} = - \frac{2}{2} = - 1

. From the parabola

x^{2} = 2 y

, we have

\frac{d y}{d x} = x

; therefore

m_{2} = 2

. Then

\tan α = \frac{(- 1) - 2}{1 + (- 1) \cdot 2} = 3.

Therefore,

α = \arctan 3 = 1.249 rad = 71.57 ° .

By symmetry, the angle of intersection at

(- 2, 2)

is the same.

Circle x^2 + y^2 = 8 and parabola x^2 = 2y intersecting at (2,2) and (-2,2)

Orthogonal Curves

Two curves are said to be orthogonal if they intersect at right angles ($90°$). The condition for this is that either slope at a point of intersection be the negative reciprocal of the other.

Show that the curves $x^{2} - y^{2} = a$ and $x y = b$ , where $a$ and $b$ are constants, intersect orthogonally.

Solution

From

x^{2} - y^{2} = a

, we obtain

2 x - 2 y \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = \frac{x}{y} .

From

x y = b

, we obtain

y + x \frac{d y}{d x} = 0 (Product Rule) ⟹ \frac{d y}{d x} = - \frac{y}{x} .

At a point of intersection,

x

and

y

have the same values in both equations. The second slope being the negative reciprocal of the first, the curves are orthogonal at all intersections.

Hyperbolas x^2 - y^2 = a and rectangular hyperbolas xy = b intersecting orthogonally

Frequently Asked Questions

What if the tangent line is vertical?

A vertical tangent occurs when f'(x_0) does not exist (or equals

\pm \infty

). In that case, the tangent line is the vertical line

x = x_{0}

, and the normal is the horizontal line

y = f (x_{0})

How do I find the tangent to an implicitly defined curve?

Use implicit differentiation to find

d y / d x

at the given point. This gives the slope of the tangent. Then substitute into the point-slope form to get the equation of the tangent line.

Can two different tangent lines from an external point have the same slope?

Yes. If the curve has two points where the tangent lines through an external point have the same slope, those tangent lines are parallel but distinct. The method of equating slopes will produce both solutions.

Tangents

Normal to a Curve

Tangent Line Through a Point Not on the Curve

Angle Between Two Curves

Frequently Asked Questions

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