Find the equation of the tangents and normal to the given curves at the indicated points
(i) y = x4 - 6x3 + 13x2 - 10x + 5 at (0, 5)
(ii) y = x4 - 6x3 + 13x2 - 10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0)
(v) x = cos t, y = sin t at t = π/4
Solution:
(i) The equation of the curve is
y = x4 - 6x3 + 13x2 - 10x + 5 at (0, 5)
On differentiating with respect to x , we get:
dy/dx = 4x3 - 18x2 + 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 = x4 - 6x3 + 13x2 - 10x + 5 at (1, 3)
On differentiating with respect to x , we get:
dy/dx = 4x3 - 18x2 + 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 = x3 at (1, 1)
On differentiating with respect to x , we get:
dy/dx = 3x2
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 = x2 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 = x4 - 6x3 + 13x2 - 10x + 5 at (0, 5) (ii) y = x4 - 6x3 + 13x2 - 10x + 5 at (1, 3) (iii) y = x3 at (1, 1) (iv) y = x2 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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