How to Find the Equation of a Tangent Line

To find the equation of a tangent line to a curve at a given point, first, find the derivative of the curve's equation, which gives the slope of the tangent. Then, use the point-slope form of a line equation, y − y₁ = m(x − x₁), where m is the slope from the derivative, and (x₁, y₁) is the point of tangency. Simplify to get the final equation.

In this article, we will answer the question "How to Find the Equation of a Tangent Line" and also learn about the tangent line.

What is a Tangent Line?

A straight line that crosses a curve at a certain point is called a tangent line. The tangent line in calculus may cross the curve in some locations and contact it in other others. The "point of tangency" is the location where the line meets the curve. One is readily seen in the example of a tangent line drawn around a circle.

A line is not a tangent line if it crosses two points on a curve without touching the curve at each location. One term for such a line is a secant line. Examples of both secant and tangent lines are shown below.

For example, PQ is a secant line because it does not touch the curve at either point P or Q; hence, it's not a tangent line.

Tangent Line Equation

The equation of the line with slope 'm' and passing through the point (x₀, y₀) is given by the point-slope form: y - y₀ = m(x - x₀). Now, let's consider the tangent line to a curve y = f(x) at the point (x₀, y₀). From what we learned earlier:

Slope of the tangent line, m = f'(x₀, y₀)

First find the equation of the tangent line. Now, substitute m, x₀ and y₀ in the point-slope form, y - y₀ = m (x - x₀).

Hence, the equation of the tangent line is:

y – y₀ = f '(x₀, y₀)(x – x₀)

Steps to Find the Equation of the Tangent Line

Find the equation of the tangent line to the curve y = f(x) at the point (x₀, y₀) or at x = x₀.

Step 1: If it's not given, that is, the question only states the tangent is at x = x₀, find the y-coordinate by plugging x₀ into the function y = f(x), i.e., y₀ = f(x₀).

Step 2: Find the derivative of the function y = f(x) and label this f'(x).

Step 3: Replace the point (x₀, y₀) into the derivative f'(x) to obtain the slope of the tangent (m).

Step 4: Use the slope-point form: y - y₀ = m(x - x₀), to write the equation of the tangent line.

Tangent Line Equation of a Parametric Curve

Sometimes, a curve isn't given in the standard form y = f(x). Instead, it might be represented using parametric equations. Let's explore how to find the tangent line equation for a parametric curve in both 2D and 3D.

Tangent Line of a Parametric Curve in 2D

If a curve in 2D is described by the parametric equations x = x(t) and y = y(t), you can find the equation of the tangent line at t = a using these steps:

Step 1: Find the point of tangency (x₀, y₀) by substituting t = a into the parametric equations.

• That is, (x₀, y₀) = (x(a), y(a)).

Step 2: Calculate the derivative using (dy/dt) / (dx/dt).

Step 3: Find the slope of the tangent (m) by substituting t = a into the derivative.

Step 4: Determine the equation of the tangent line using the formula y - y₀ = m(x - x₀).

Tangent Line of a Parametric Curve in 3D

If a curve in 3D is described by the parametric equations x = x(t), y = y(t), and z = z(t), follow these steps to find the tangent line equation at t = t₀:

Step 1: Find the point of tangency (x₀, y₀, z₀) by substituting t = t₀ into the parametric equations.

That is, (x₀, y₀, z₀) = (x(t₀), y(t₀), z(t₀)).

Step 2: Find the derivatives x'(t), y'(t), and z'(t).

Step 3: Calculate the direction ratios <a, b, c> by substituting t = t₀ into these derivatives.

That is, <a, b, c> = <x'(t₀), y'(t₀), z'(t₀)>.

Step 4: Write the equation of the tangent line using one of the following formulas:

x = x₀ + at, y = y₀ + bt, z = z₀ + ct, or

⇒ (x - x₀) / a = (y - y₀) / b = (z - z₀) / c.

Tangent Line of a Polar Curve

When a function is defined by a polar equation r = r(t), you can find the equation of the tangent line at t = a by following these steps:

Step 1: Find 'r' using r = r(a).

Step 2: Determine the point (x₀, y₀) where the tangent is drawn using (x₀, y₀) = (r cos(a), r sin(a)).

Step 3: Calculate dy/dx with the formula:

Step 4: Find the slope of the tangent (m) by substituting t = a into dy/dx.

Step 5: Write the equation of the tangent line using the formula y - y₀ = m(x - x₀).

Important Notes on Tangent Lines

• The equation of the tangent line to the curve y = f(x) at the point (x₀, y₀) is y – y₀ = m(x – x₀), where m = f ̋(x) at (x₀, y₀).
• If θ is the angle the tangent line makes with the positive x-axis, its slope is m =.
• The normal line to the curve at a point, and the tangent line to the curve at the given point, are perpendicular to each other. So then, if \text{slope of the tangent is } m, \text{ then } \text{ slope of the normal } = - \frac{1}{m}.
• A curve y = f(x) has horizontal tangents at points where f'(x) = 0 because horizontal tangents are parallel to the x-axis.
• A curve y = f(x) has vertical tangents where the function f'(x) is undefined because the vertical tangents run parallel to the y-axis.
• The equation of the tangent line is used to find approximate values of a function near the point at which the tangent is drawn.

Also Read,

• Tangent and Normal
• Equation of Tangents and Normals
• Application of Derivative
• How to Find the Slope of a Tangent Line?

Examples on Tangent Line Equation

Example 1: Find the equation of the tangent line of the curve y = 3x² - 4x at x = -1. Also, verify it.

Solution:

The point at which the tangent is drawn is, (x₀, y₀) = (-1, f(-1)) = (-1, 3(-1)² - 4(-1)) = (-1, 7).

The given curve is, f(x) = 3x² - 4x.

Its derivative is f'(x) = 6x - 4.

The slope of the tangent is, m = f'(-1) = 6(-1) - 4 = -10.

The equation of the tangent line is,

y - y₀ = m (x - x₀)

⇒ y - 7 = -10 (x - (-1))

⇒ y - 7 = -10 (x + 1)

⇒ y - 7 = -10x - 10

⇒ y = -10x - 3

Example 2: What is the tangent line equation of the curve x = cos t and y = sin t at t = π/2?

Solution:

The point at which the tangent is drawn is, (x₀, y₀) = (cos π/2, sin π/2) = (0, 1). (by using unit circle)

The curve is defined by parametric equations.

So dy/dx = (dy/dt) / (dx/dt) = (cos t) / (- sin t).

The slope of the tangent is, m = (dy/dx)ₜ ₌ π/₂ = (cos π/2) / (-sin π/2) = 0/(-1) = 0.

The equation of the tangent line is,

y - y₀ = m (x - x₀)

⇒ y - 1 = 0 (x - 0)

⇒ y - 1 = 0

⇒ y = 1

Example 3: Find the parametric equations of the tangent line drawn to the curve given by the equations x = 2t², y = 3t, z = t³ at t = 2.

Solution:

The point at which the tangent line is drawn is,

(x₀, y₀, z₀) = (x(2), y(2), z(2)) = (2(2)², 3(2), 2³) = (8, 6, 8).

Compute the derivatives

x'(t) = 4t, y'(t) = 3, z'(t) = 3t²

The direction ratios of the tangent line are,

<a, b, c> = <x'(2), y'(2), z'(2)> = <4(2), 3, 3(2)²> = <8, 3, 12>

The parametric equations of the tangent line are:

x = x₀ + at, y = y₀ + bt, z = z₀ + ct

⇒ x = 8 + 8t, y = 6 + 3t, z = 8 + 12t