# Find the equation of the tangents and normal to the given curves at the indicated points

(i) y = x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (0, 5)

(ii) y = x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (1, 3)

(iii) y = x^{3} at (1, 1)

(iv) y = x^{2 }at (0, 0)

(v) x = cos t, y = sin t at t = π/4

**Solution:**

(i) The equation of the curve is

y = x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (0, 5)

On differentiating with respect to x , we get:

dy/dx = 4x^{3} - 18x^{2} + 26x - 10

dy/dx]_{(0, 5)} = - 10

Thus,

the slope of the tangent at (0, 5) is - 10.

The equation of the tangent is given as:

y - 5 = - 10( x - 0)

⇒ y - 5 = - 10x

⇒ 10x + y = 5

Slope of normal at (0, 5) is

- 1/slope of the tangent at (0, 5)

= -1/- 10

= 1/10

Therefore, the equation of the normal at (0, 5) is given as:

y - 5 = 1/10 (x - 0)

⇒ 10 y - 50 = x

⇒ x - 10 y + 50 = 0

(ii) The equation of the curve is

y = x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (1, 3)

On differentiating with respect to x , we get:

dy/dx = 4x^{3} - 18x^{2} + 26x - 10

dy/dx]_{(1 - 3)} = = 4 - 18 + 26 - 10 = 2

Thus,

the slope of the tangent at (1, 3) is 2.

The equation of the tangent is given as:

y - 3 = 2 (x - 1)

⇒ y - 3 = 2x - 2

⇒ y = 2x + 1

Slope of normal at (1, 3) is

- 1/slope of the tangent at (1, 3) = - 1/2

Therefore,

the equation of the normal at (1, 3) is given as:

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

⇒ 2 y - 6 = x + 1

⇒ x + 2y - 7 = 0

(iii) The equation of the curve is

y = x^{3} at (1, 1)

On differentiating with respect to x , we get:

dy/dx = 3x^{2}

dy/dx]_{(1, 1)} = 3(1)^{2} = 3

Thus, the slope of the tangent at (1, 1) is 3.

The equation of the tangent is given as:

⇒ y - 1 = 3(x - 1)

⇒ y = 3x - 2

Slope of normal at (1, 1) is

- 1/slope of the tangent at (1, 1) = - 1/3

Therefore,

the equation of the normal at (1,1) is given as:

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

⇒ 3y - 3 = - x + 1

⇒ x + 3y - 4 = 0

(iv) The equation of the curve is y = x^{2 }at (0, 0)

On differentiating with respect to x , we get:

dy/dx = 2x

dy/dx]_{(0, 0)} = 0

Thus,

the slope of the tangent at (0, 0) is 0.

The equation of the tangent is given as:

y - 0 = 0 ( x - 0)

⇒ y = 0

Slope of normal at (0, 0) is

- 1/slope of the tangent at (0, 0) = - 1/0, which is not defined.

Therefore,

the equation of the normal at (0, 0) is given as x = 0

(v) The equation of the curve is

x = cos t, y = sin t at t = π/4

On differentiating with respect to t, we get:

dx/dt = - sin t and dy/dt = cos t

Therefore,

dy/dx = (dy/dt)/(dx/dt)

= (cos t)/(- sin t)

= - cot t

Hence,

dy/dx]_{t = π/4} = - cot t = - 1

Thus,

the slope of the tangent at t = π / 4 is - 1.

Hence,

x = 1/√2 and y = 1/√2

The equation of the tangent is given as:

⇒ y - 1/√2 = - 1 (x - 1/√2)

⇒ x + y - 1/√2 - 1/√2 = 0

⇒ x + y - √2 = 0

Slope of normal at t = π / 4 is

- 1 / slope of the tangent at t = π / 4

= (- 1)/(- 1) = 1

Therefore, the equation of the normal at t = π / 4 is given as:

⇒ y - 1/√2

= 1 (x - 1/√2)

⇒ x = y

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 14

## Find the equation of the tangents and normal to the given curves at the indicated points (i) y = x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (0, 5) (ii) y = x^{4} - 6x^{3} + 13x^{2} - 10x + 5 at (1, 3) (iii) y = x^{3} at (1, 1) (iv) y = x^{2 }at (0, 0) (v) x = cos t, y = sin t at t = π/4.

**Summary:**

The equation of the tangents and normal to the given curves at the indicated points are as follows: x - 10 y + 50 = 0, x + 2y - 7 = 0, x + 3y - 4 = 0, x = 0 and x = y respectively

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