# Tangent - Equation Of Tangent And Normal

Tangent

Tangent is a straight line which touches the curve at a given point and it has the same slope of the curve at that point.

Consider the function y = f(x) given in the figure,

First derivative of a function at a point gives its slope of the tangent at that point.

For the function f, first derivative i.e. f’at the point where x =a  is f‘(a).

f‘(a) = slope of the tangent at the point where x =a

Equation of a straight line $\frac {(y-y_1)}{(x-x_1)}$ = m , where “m” is slope of the line.

$\frac{(y – f(a))}{(x-a)}$= f’(a) Is the equation of tangent of the function at x=a.

For example,

Consider the function f(x) = $x^3 – 3x^2 + 2x + 1$ , . Find equation of tangent of f(x) when . x = 2.

$~~~~~~~~~~~~~~~~~~~~~~~~~$ f(x) = $x^3 – 3x^2 + 2x + 1$

$~~~~~~~~~~~~~~~~~~~~~~~~~$ f(2) = $2^3 – 3 × 2^2 + 2 × 2 + 1 = 8 – 12 + 4 + 1 = 1$

First derivative of the function f(x).f‘(x) = $\frac {d}{dx} ( x^3 – 3x^2 + 2x + 1 )$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ = $3x^2 – 6x + 2$

f‘(2)= 3 × 22 – 6 × 2 + 2 = 12 – 12 + 2 = 2

Equation of the tangent = $\frac {(y-1)}{(x-2)}$ = 2.

y – 1 = 2(x – 2)

y – 1 = 2x – 4

y = 2x – 3 is the equation of tangent.

Normal to a curve:

Normal is a line which is perpendicular to the tangent. Tangent and normal of f(x) is drawn in the figure below.

Since tangent and normal are perpendicular to each other, product of slope of the tangent and slope of the normal will be equal to -1.

$\frac{(y – f(a))}{(x-a)}$ = f‘(a)Is the equation of tangent of the function y = f(x) at x = a .

Then, equation of the normal will be,$\frac{(y – f(a))}{(x-a)}$ = $\frac {-1}{f'(a)}$

For example,

Consider the function,f(x) = $x^2 – 2x + 5$ . Find the equation of tangent and equation of normal at x = 3.

f(x) = $x^1 – 2x + 5$

f(x) = 32 – 2 × 3 + 5 = 9 – 6 + 5 = 8

First derivative, f‘(x) = $\frac{d}{dx}(x^2-2x+5)$ = 2x – 2

f(‘x) when x = 3,f(3) = 2 × 3 – 2 = 6 – 2 = 4

Equation of the tangent is,$\frac{(y-8)}{(x-3)}$=  4

y – 8 = 4(x – 3)

y – 8 = 4x – 12

y = 4x – 4, is equation of tangent.

Slope of normal = – $\frac{1}{slope~of~tangent}$ = – $\frac 14$

Equation of normal is,$\frac{(y-8)}{(x-3)}$= –$\frac{1}{4}$

$~~~~~~~~~~~~$ 4(y – 8) = (3 – x)

$~~~~~~~~~~~~$ 4y – 32 = 3 – x

$~~~~~~~~~~~~$ y = –$\frac x4$ + $\frac {35}{4}$<

You are now good with the concept of tangents and normals, to know about tangents of a circle, log on to www.byjus.com and learn new concepts every day!

#### Practise This Question

The  two curves x33xy2+2=0 and 3x2yy32=0