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,

Tangent

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.

Tangent - Normal to a curve

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

For the curve xy=c2 the subnormal at any point varies as
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