## Tangent and Normal Question1

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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