EQUATION OF TANGENT AND NORMAL TO THE CURVE

Tangent is also a line which touches the curve. To find the equation of tangent, we have to follow the given below.

i) Find the slope of the tangent drawn at the point (x1, y1) from the given equation of curve.

Use the formula,

y - y1 = m(x - x1)

ii)  Here m is the slope of the tangent line at the point of contact. (x1, y1) is the point of contact.

tangent-and-normal-to-the-curve

Normal is line the perpendicular to tangent line drawn at the point of contact.

We use the formula 

y - y1 = m(x - x1)

Find the equation of tangent and normal to the curve at the given point.

Problem 1 :

y = x – 2x2 + 3 at x = 2

Solution :

y = x – 2x2 + 3 at x = 2

When, x = 2, then

y = 2 - 2(2)2 + 3

= 2 - 8 + 3

= -6 + 3

y = -3

dy/dx = 1 - 4x

= 1 - 4(2)

= 1 - 8

= -7

So, the required point is (2, -3).

Equation of tangent :

(y - y1) = m(x - x1)

(y + 3) = -7(x - 2)

y + 3  = -7(x - 2)

y + 3 = -7x + 14

7x + y + 3 - 14 = 0

7x + y - 11 = 0

Equation of normal :

(y - y1) = (-1/m)(x - x1)

(y + 3) = 1/7(x - 2)

7(y + 3) = 1(x - 2)

7y + 21 = x - 2

x - 2 - 7y - 21 = 0

x - 7y - 23 = 0

Problem 2 :

y = √x + 1 at x = 4

Solution :

y = √x + 1 at x = 4

When, x = 4, then 

y = √4 + 1 

= 2 + 1

y = 3

dy/dx = 1/2√x

= 1/2√4

= 1/2(2)

= 1/4

So, the required point is (4, 3).

Equation of tangent :

(y - y1) = m(x - x1)

(y - 3) = 1/4(x - 4)

4(y - 3)  = x - 4

4y - 12 = x - 4

x - 4y + 12 - 4 = 0

x - 4y + 8 = 0

Equation of normal :

(y - y1) = (-1/m)(x - x1)

(y - 3) = -4(x - 4)

y - 3 = -4x + 16

4x + y - 3 - 16 = 0

4x + y - 19 = 0

Problem 3 :

y = x3 - 5x at x = 1

Solution :

y = x3 - 5x at x = 1

When, x = 1, then 

y = (1)3 - 5(1)

= 1 - 5

y = -4

dy/dx = 3x2 - 5

= 3(1) - 5

= 3 - 5

= -2

So, the required point is (1, -4).

Equation of tangent :

(y - y1) = m(x - x1)

(y + 4) = -2(x - 1)

y + 4  = -2x + 2

2x + y + 4 - 2 = 0

2x + y + 2 = 0

Equation of normal :

(y - y1) = (-1/m)(x - x1)

(y + 4) = (-1/-2)(x - 1)

(y + 4) = (1/2)(x - 1)

2(y + 4) = x - 1

2y + 8 = x - 1

x - 2y - 8 - 1

x - 2y - 9 = 0

Problem 4 :

y = 4/√x at (1, 4)

Solution :

y = 4/√x at (1, 4)

dy/dx = -2/(√x)3

= -2/(√1)3

= -2/√1

= -2/1

Slope (m) = -2

So, the required point is (1, 4).

Equation of tangent :

(y - y1) = m(x - x1)

(y - 4) = -2(x - 1)

y - 4  = -2x + 2

2x + y - 4 - 2 = 0

2x + y - 6 = 0

Equation of normal :

(y - y1) = (-1/m)(x - x1)

(y - 4) = (-1/-2)(x - 1)

(y - 4) = (1/2)(x - 1)

2(y - 4) = 1(x - 1)

2y - 8 = x - 1

x - 2y + 8 - 1 = 0

x - 2y + 7 = 0

Problem 5 :

y = 3x - 1x2 at (-1, -4)

Solution :

y = 3x - 1x2dydx = -3x2 + 2x3= -3(-1)2 + 2(-1)3= -31 + 2-1= -3 - 2= -5

So, the required point is (-1, -4).

Equation of tangent :

(y - y1) = m(x - x1)

(y + 4) = -5(x + 1)

y + 4  = -5x - 5

5x + y + 4 + 5 = 0

5x + y + 9 = 0

Equation of normal :

(y - y1) = (-1/m)(x - x1)

(y + 4) = (-1/-5)(x - 1)

(y + 4) = (1/5)(x - 1)

5(y + 4) = 1(x - 1)

5y + 20 = x - 1

x - 5y - 20 - 1 = 0

x - 5y - 21 = 0

Problem 6 :

y = 3x2 - 1/x at x = -1.

Solution :

y = 3x2 - 1/x at x = -1.

When, x = -1

y = 3(-1)2 - 1/(-1)

= 3 + 1

= 4

dy/dx = 6x + 1/x2

= 6(-1) + 1/(-1)2

= -6 + 1

= -5

So, the required point is (-1, 4).

Equation of tangent :

(y - y1) = m(x - x1)

(y - 4) = -5(x + 1)

y - 4  = -5x - 5

5x + y - 4 + 5 = 0

5x + y + 1 = 0

Equation of normal :

(y - y1) = (-1/m) (x - x1)

(y - 4) = (-1/-5)(x + 1)

-5(y - 4) = -1(x + 1)

-5y + 20 = -x - 1

x - 5y + 20 + 1 = 0

x - 5y + 21 = 0

Recent Articles

  1. Solving Word Problems on Numbers Worksheet

    Feb 25, 24 07:44 AM

    Solving Word Problems on Numbers Worksheet

    Read More

  2. Solving Word Problems on Numbers

    Feb 24, 24 11:07 PM

    Solving Word Problems on Numbers

    Read More

  3. Reducing Ratios to Lowest Terms Worksheet

    Feb 24, 24 08:49 PM

    Reducing Ratios to Lowest Terms Worksheet

    Read More