Tangent & Normal

Tangents and normals are the lines associated with curves. A line touching the curve at a distinct point is called a tangent. Each of the points on the curve has a tangent. A line perpendicular to the tangent at the point of contact is called a normal.

For a curve given by y = f(x), \( \frac{dy}{dx} \) is the gradient function. If given the coordinates of a point as on the curve, The value of \( \frac{dy}{dx} \) and the equation of the tangent at the point can be found.

Note: if two lines with gradients m₁, m₂ are perpendicular,
then m₁ × m₂ = -1

Hence the gradient of the line perpendicular to the tangent, the normal to the curve is \( \scriptsize m_2 = \normalsize \frac{-1}{m_1} \)

m₂ = \( \frac {-1}{\large \frac{dy}{dx}} \)

Example 4.1.1:

Find the equation of the tangent and normal to the curve y = (1 – x)(2 + x), at the point where x = 2.

Solution:

y = (1 – x)(2 + x)

y = 2 – x – x²

⇒ \( \frac{dy}{dx} \) = -1 – 2x

x = 2

y = 2 – 2 – (2²)

y = -4

Gradient of the tangent at the point (2, -4)

Equation of line through (x, y₁) with gradient m is:

y – y₁ = m(x – x₁)

∴ y – (-4) = -5(x – 2)

⇒ y + 4 = -5(x – 2)

⇒ y + 4 = -5x + 10

⇒ y + 5x = 10 – 4

So its equation is y + 5x = 6

Gradient of the normal at the point = \(\scriptsize m_2 = \normalsize \frac{-1}{m_1} \\ = \frac {-1}{\large \frac{dy}{dx}} \\= \frac{-1}{-5} \\= \frac{1}{5} \)

and its equation is ⇒ y + 4 = \( \frac {1}{5} \scriptsize(x \; – \; 2) \\ \scriptsize 5y = x \; – \; 22\)