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Equation of Tangent And Normal to a Curve with Examples, Class 12 Math Notes Study Material Download Free PDF

 The applications of derivatives are:

  • determining the rate of change of quantities
  • finding the equations of tangent and normal to a curve at a point
  • finding turning points on the graph of a function which in turn will help us to locate points at which largest or smallest value (locally) of a function occurs.

In this article, we will learn to use differentiation to find the equation of the tangent line and the normal line to a curve at a given point.

Tangent and Normal Equation

Tangent and Normal Equation

We know that the equation of the straight line that passes through the point (x0, y0) with finite slope “m” is given as

y – y0 = m (x – x0)

It is noted that the slope of the tangent line to the curve f(x)=y at the point (x0, y0) is given by

dydx](x0,y0)(=f′(x0))

Therefore, the equation of the tangent (x0, y0) to the curve y=f(x) is 

y – y0 = f ′(x0)(x – x0)

Also, we know that normal is the perpendicular to the tangent line. Hence, the slope of the normal to the curve f(x)=y at the point (x0, y0) is given by -1/f’(x0), if f’(x0) ≠ 0.

Hence, the equation of the normal to the curve y=f(x) at the point (x0, y0) is given as:

y-y0 = [-1/f’(x0)] (x-x0)

The above expression can also be written as

(y-y0) f’(x0) + (x-x0) = 0

Points to Remember

  • If a tangent line to the curve y = f (x) makes an angle θ with x-axis in the positive direction, then dy/dx = slope of the tangent = tan = Î¸.
  • If the slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x0, y0) is given by y = y0
  • If θ →Ï€/2, then tan θ → ∞, which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at (x0, y0) is given by x = x0

Equation of Tangent and Normal Problems

Go through the below tangent and normal problems:

Example 1:

Find the equation of a tangent to the curve y = (x-7)/[(x-2)(x-3)] at the point where it cuts the x-axis.

Solution:

As the point cut at the x-axis, then y=0. Hence, the equation of the curve, if y=0, then the value of x is 7. (i.e., x=7). Hence, the curve cuts the x-axis at (7,0)

Now, differentiate the equation of the curve with respect to x, we get

dy/dx = [(1-y)(2x-5)] / [(x-2)((x-3)]

dy/dx](7, 0) = (1-0)/[(5)(4)] = 1/20

Hence, the slope of the tangent line at (7, 0) is 1/20.

Therefore, the equation of the tangent at (7, 0) is 

Y-0 = (1/20)(x-7)

20y-x+7 = 0.

Example 2: 

Find the equation of tangent and normal to the curve x(⅔)+ y(⅔) = 2 at (1, 1)

Solution:

Given curve: x(⅔)+ y(⅔) = 2

Finding Equation of Tangent:

Now, differentiate the curve with respect to x, we get

(⅔)x(-⅓) + (⅔)y(-⅓) dy/dx = 0

The above equation can be written as:

dy/dx = -[y/x]⅓

Hence, the slope of the tangent at the point (1, 1) is dy/dx](1,1) = -1

Now, substituting the slope value in the tangent equation, we get

Equation of tangent at (1, 1) is 

y-1 = -1(x-1)

y+x-2 = 0

Thus, the equation of tangent to the curve at (1, 1) is y+x-2 =0

Finding Equation of Normal:

The slope of the normal at the point (1, 1) is 

= -1/slope of the tangent at (1, 1)

= -1/ -1

=1

Therefore, the slope of the normal is 1. 

Hence, the equation of the normal is 

y-1 = 1(x-1)

y-x = 0

Therefore, the equation of the normal to the curve at (1, 1) is y-x =0

Practice problems

Solve the following problems:

  1. Calculate the slope of the tangent to the curve y=x3 -x at x=2.
  2. Determine the slope of the tangent to the curve y=x3-3x+2 at the point whose x-coordinate is 3.
  3. Find the equation of tangent and normal to the curve y = x3 at (1, 1).
  4. Find the equation of normal at the point (am2, am3) for the curve ay2=x3.
  5. The slope of the normal to the curve y=2x2 + 3 sin x at x=0 is

               (a)3  (b)  -3 (c)⅓  (d) -⅓

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