EQUATION OF NORMAL TO THE CURVE WITH DERIVATIVE

Step 1 :

By finding the first derivative, we get slope of the tangent line drawn to the curve.

Step 2 :

By applying the specific point in the general slope, we can find slope of the tangent line drawn at the specific point.

Step 3 :

From slope of tangent we have to find the slope of normal (-1/m).

Step 4 :

Now we have to apply the point and the slope in the formula

(y - y₁)  =  (-1/m) (x - x₁)

Example 1 :

Find the equation of the tangent to the curve

y  =  x³

at the point (1, 1)

Solution :

y  =  x³

Step 1 :

dy/dx  =  3x²

slope of the curve at the point (1, 1)

dy/dx  =  3(1)²

Slope at (1, 1) :

dy/dx  =  3

Slope of normal  =  -1/3

Equation of the normal :

(y - y₁)  =  (-1/m) (x - x₁)

(y-1)  =  (-1/3) (x-1)

3(y-1)  =  -1(x-1)

3y-3  =  -x+1

x+3y-3-1  =  0

x+3y-4  =  0

The required equation of normal is x+3y-4  =  0.

Example 2 :

Find the equation of normal to the curve

y  =  x²-x-2

at the point (1, -2)

Solution :

y  =  x²-x-2

dy/dx  =  2x-1

Slope at the point (1, -2) :

dy/dx  =  2(1) - 1

= 2-1

dy/dx  =  1

slope of the tangent (dy/dx) = 1

Slope of normal  =  -1/1

=  -1

Equation of the normal :

(y - y₁)  =  (-1/m) (x - x₁)

(y-(-2))  =  -1 (x-1)

y+2  =  -x+1

x+y+2-1  =  0

x+y+1  =  0

Equation of normal is x+y+1  =  0.

Example 3 :

Find a point on the curve y = √x, where the tangent makes an angle of 45 degrees with the positive x-axis.

Solution :

y  =  √x

m  =  tan θ

m  =  tan 45

m  =  1 ---(1)

dy/dx  =  1/2√x  ---(2)

(1)  =  (2)

1  =  1/2√x

2√x  =  1

4x  =  1

x  =  1/4

By applying the value of x in y  =  √x, we get

y  =  √(1/4)

y   =  ± 1/2

Since the tangent drawn at the positive x-axis, we find equation of the tangent at the point (1/4, 1/2).

Equation of the tangent line :

y-y₁  =  m(x-x₁)

y-(1/2)  =  1(x-1/4)

(2y-1)/2  =  1(4x-1)/4

4(2y-1)  =  2(4x-1)

8y-4  =  8x-2

8x-8y-2+4  =  0

8x-8y+2  =  0

4x-4y+1  =  0

So, equation of tangent is 4x-4y+1  =  0.

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