# 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.

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 + 3y = -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 + 1y = 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 - 5y = -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 :

Solution :

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

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