Tangents and normals

The derivative $f'(a)$ is the slope of the line tangent to the graph of $f$ at the point $(a, f(a))$. The normal at the same point is the line perpendicular to the tangent. These two lines reveal local geometry , how the curve bends and how a normal would push away from it.

Equation of the tangent

The line tangent to $y = f(x)$ at $(a, f(a))$:

$y - f(a) = f'(a)(x - a).$

Equation of the normal

The normal is perpendicular to the tangent. If the tangent has slope $m$, the normal has slope $-1/m$:

$y - f(a) = -\frac{1}{f'(a)}(x - a), \quad \text{provided } f'(a) \neq 0.$

If $f'(a) = 0$, the tangent is horizontal and the normal is the vertical line $x = a$.

A worked example

For $y = x^2$ at the point $(2, 4)$:

$f'(x) = 2x$, $f'(2) = 4$. Tangent: $y - 4 = 4(x - 2)$, i.e., $y = 4x - 4$.

Normal: $y - 4 = -\tfrac{1}{4}(x - 2)$, i.e., $y = -\tfrac{1}{4}x + \tfrac{9}{2}$.

Tangent and normal in different forms

For a curve given parametrically by $x = f(t), y = g(t)$, the slope of the tangent at parameter $t$ is $\dfrac{g'(t)}{f'(t)}$.

For an implicit curve $F(x, y) = 0$, differentiate implicitly to get $\dfrac{dy}{dx}$ at the point of interest.

Length of tangent / normal / subtangent / subnormal

At a point $P = (x_0, y_0)$ with $\dfrac{dy}{dx} = m$:

• Subtangent length: $\big|\dfrac{y_0}{m}\big|$.
• Subnormal length: $|m y_0|$.
• Tangent length: $|y_0|\sqrt{1 + 1/m^2}$.
• Normal length: $|y_0|\sqrt{1 + m^2}$.

These appear in older problems and in some applications to geometry.

Worked examples

Example 1. Find the equation of the tangent to $y = x^3 - 3x + 5$ at $x = 1$.

$y(1) = 1 - 3 + 5 = 3$. $y'(x) = 3x^2 - 3$, $y'(1) = 0$. So tangent is horizontal: $y = 3$.

Example 2. Find tangent and normal to $y = \ln x$ at $(1, 0)$.

$y'(x) = 1/x$, $y'(1) = 1$. Tangent: $y = x - 1$. Normal: $y = -(x - 1) = 1 - x$.

Example 3. Find the points on $y = x^3$ at which the tangent is parallel to $y = 3x + 1$.

Need $y'(x) = 3x^2 = 3$, i.e., $x = \pm 1$. Points: $(1, 1)$ and $(-1, -1)$.

Example 4. Find the tangent to $x^2 + y^2 = 25$ at $(3, 4)$.

Implicit: $2x + 2y \dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -x/y = -3/4$.

Tangent: $y - 4 = -\tfrac{3}{4}(x - 3)$, i.e., $3x + 4y = 25$. (Standard tangent equation to a circle.)

Example 5. Find the angle between the tangents to $y = x^2$ at $(1, 1)$ and at $(-1, 1)$.

Slopes: $2$ and $-2$. The angle $\theta$ between two lines of slopes $m_1, m_2$ satisfies $\tan\theta = \big|\dfrac{m_1 - m_2}{1 + m_1 m_2}\big| = \big|\dfrac{4}{-3}\big| = 4/3$.

So $\theta = \tan^{-1}(4/3)$.

Example 6. Find the equation of the normal to the curve $y = x^2 + 2x + 1$ which passes through $(1, 4)$.

Let the normal be at $(a, a^2 + 2a + 1)$. Slope of tangent there: $2a + 2$. Slope of normal: $-1/(2a + 2)$.

Normal equation: $y - (a^2 + 2a + 1) = -\dfrac{1}{2a + 2}(x - a)$. Passes through $(1, 4)$:

$4 - a^2 - 2a - 1 = -\dfrac{1 - a}{2a + 2}$.

$3 - a^2 - 2a = -\dfrac{1 - a}{2(a + 1)} = \dfrac{a - 1}{2(a + 1)}$.

Multiply: $2(a + 1)(3 - a^2 - 2a) = a - 1$. Solving this cubic gives the points of tangency for the normals passing through $(1, 4)$.

Try it yourself

1. Find the tangent and normal to $y = x^3 - x$ at $x = 1$.
2. Find points on $y = x^2 - 4x + 5$ where tangent is horizontal.
3. Find tangent to $x^2 + y^2 = 1$ at the point $(\cos\theta, \sin\theta)$.
4. At what point on $y = \sin x$ is the tangent slope equal to $1/2$?
5. Find equation of tangent at $(1, 1)$ to $y = \dfrac{1}{x^2 - 3x + 3}$.
6. Find the equation of the tangent at $(a\cos^3 t, a\sin^3 t)$ on the astroid $x^{2/3} + y^{2/3} = a^{2/3}$.
7. Find the points where the tangent to $y = x^3 - 11 x + 5$ has slope $1$.
8. Find the angle between the tangents to $y^2 = 4x$ at the points $(1, 2)$ and $(1, -2)$.
9. Show that the tangent to $y = e^x$ at $(0, 1)$ has equation $y = 1 + x$.
10. Find the equation of the normal to $xy = c^2$ at $(c, c)$.
11. A curve $y = f(x)$ has $f(1) = 2$ and $f'(1) = -3$. Write tangent and normal at $(1, 2)$.
12. Find the length of the subtangent at any point on the curve $y^2 = 4ax$.
13. Find tangent to $y = (\log x)^2$ at $x = 1$.
14. At what angle does $y = x^3 - 3x$ cross the $x$-axis at $x = 0$?

Pitfalls / Tricks

• The slope of the normal is $-1/m$ (negative reciprocal), not just $-m$.
• A horizontal tangent gives a vertical normal: $x = a$.
• When asked for a tangent from a point not on the curve, set up the point of tangency as unknown.
• For implicit curves, differentiate implicitly , do not try to solve for $y$ first.
• Always plug the value of $x = a$ into $y'$ to get a number, before writing the line equation.

Next, approximations using differentials.