Tangent and Normal Question1





In this page tangent and normal question1 we are going to see solution of first question.

(1) Find the equation of the tangent and normal of the following curves

(i) y = x² - 4 x - 5, at x = -2

To find the equation of the tangent line we need to have two information that is a point on the tangent and the slope of the tangent line.

Here we have a point, to find the slope of the tangent line at that particular point we have to differentiate the given equation

y = x² - 4 x - 5

dy/dx = 2 x - 4

        = 2 (-2) - 4

        = -4 - 4

        = -8

substitute x = -2 in the given equation to get y-coordinate value

y = (-2)² - 4 (-2) - 5

   = 4 + 8 - 5

   = 12 - 5

   = 7

So the required point is (-2,7)

Equation of tangent :

(y - y₁) = m (x - x₁)

(y - 7) = -8 (x - (-2))

 y - 7 = -8 (x + 2)

 y - 7 = -8 x - 16

8 x + y -7 + 16 = 0

8 x + y + 9 = 0

Equation of normal :

(y - y₁) = (-1/m) (x - x₁)

(y - 7) = (1/8) (x - (-2))

 8(y - 7) = 1 (x + 2)

8 y - 56 = x + 2

x - 8 y + 2 + 56 = 0

x - 8 y + 58 = 0


(ii) y = x - sin x cos x, at x = π/2

To find the equation of the tangent line we need to have two information that is a point on the tangent and the slope of the tangent line.

Here we have a point, to find the slope of the tangent line at that particular point we have to differentiate the given equation

y = x - sin x cos x

dy/dx = 1 - [sin x (- sin x) + cos x (cos x)]

         = 1 - [- sin² x + cos² x]

         = 1 - [cos² x - sin² x]

         = 1 - cos 2 x

slope of the tangent at x = π/2

         = 1 - cos 2 (π/2)

         = 1 - cos π

         = 1 - (-1)

         = 2 

substitute x = π/2 in the given equation to get y-coordinate value

y = x - sin x cos x

   = π/2 - sin π/2 cos π/2

   = π/2 - (1) (0)

   = π/2 

So the required point is (π/2,π/2)

Equation of tangent :

(y - y₁) = m (x - x₁)

[y - (π/2)] = 2 [x - (π/2)]

[y - (π/2)] = 2 x - π

2 x - y - π + (π/2) = 0

2 x - y - (π/2) = 0

Equation of normal :

(y - y₁) = (-1/m) (x - x₁)

[y - (π/2)] = (-1/2) [x - (π/2)]

2[y - (π/2)] = -1[x - (π/2)]

2 y - π  = - x + (π/2)

x + 2 y - π - (π/2) = 0

x + 2 y - (3π/2) = 0