Tangent is also a line which touches the curve. To find the equation of tangent, we have to follow the given below.
i) Find the slope of the tangent drawn at the point (x_{1}, y_{1}) from the given equation of curve.
Use the formula,
y - y_{1} = m(x - x_{1})
ii) Here m is the slope of the tangent line at the point of contact. (x_{1}, y_{1}) is the point of contact.
Normal is line the perpendicular to tangent line drawn at the point of contact.
We use the formula
y - y_{1} = m(x - x_{1})
Find the equation of tangent and normal to the curve at the given point.
Problem 1 :
y = x – 2x^{2} + 3 at x = 2
Solution :
y = x – 2x^{2} + 3 at x = 2
When, x = 2, then
y = 2 - 2(2)^{2} + 3 = 2 - 8 + 3 = -6 + 3 y = -3 |
dy/dx = 1 - 4x = 1 - 4(2) = 1 - 8 = -7 |
So, the required point is (2, -3).
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y + 3) = -7(x - 2)
y + 3 = -7(x - 2)
y + 3 = -7x + 14
7x + y + 3 - 14 = 0
7x + y - 11 = 0
Equation of normal :
(y - y_{1}) = (-1/m)(x - x_{1})
(y + 3) = 1/7(x - 2)
7(y + 3) = 1(x - 2)
7y + 21 = x - 2
x - 2 - 7y - 21 = 0
x - 7y - 23 = 0
Problem 2 :
y = √x + 1 at x = 4
Solution :
y = √x + 1 at x = 4
When, x = 4, then
y = √4 + 1 = 2 + 1 y = 3 |
dy/dx = 1/2√x = 1/2√4 = 1/2(2) = 1/4 |
So, the required point is (4, 3).
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - 3) = 1/4(x - 4)
4(y - 3) = x - 4
4y - 12 = x - 4
x - 4y + 12 - 4 = 0
x - 4y + 8 = 0
Equation of normal :
(y - y_{1}) = (-1/m)(x - x_{1})
(y - 3) = -4(x - 4)
y - 3 = -4x + 16
4x + y - 3 - 16 = 0
4x + y - 19 = 0
Problem 3 :
y = x^{3} - 5x at x = 1
Solution :
y = x^{3} - 5x at x = 1
When, x = 1, then
y = (1)^{3} - 5(1) = 1 - 5 y = -4 |
dy/dx = 3x^{2} - 5 = 3(1) - 5 = 3 - 5 = -2 |
So, the required point is (1, -4).
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y + 4) = -2(x - 1)
y + 4 = -2x + 2
2x + y + 4 - 2 = 0
2x + y + 2 = 0
Equation of normal :
(y - y_{1}) = (-1/m)(x - x_{1})
(y + 4) = (-1/-2)(x - 1)
(y + 4) = (1/2)(x - 1)
2(y + 4) = x - 1
2y + 8 = x - 1
x - 2y - 8 - 1
x - 2y - 9 = 0
Problem 4 :
y = 4/√x at (1, 4)
Solution :
y = 4/√x at (1, 4)
dy/dx = -2/(√x)^{3}
= -2/(√1)^{3}
= -2/√1
= -2/1
Slope (m) = -2
So, the required point is (1, 4).
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - 4) = -2(x - 1)
y - 4 = -2x + 2
2x + y - 4 - 2 = 0
2x + y - 6 = 0
Equation of normal :
(y - y_{1}) = (-1/m)(x - x_{1})
(y - 4) = (-1/-2)(x - 1)
(y - 4) = (1/2)(x - 1)
2(y - 4) = 1(x - 1)
2y - 8 = x - 1
x - 2y + 8 - 1 = 0
x - 2y + 7 = 0
Problem 5 :
Solution :
So, the required point is (-1, -4).
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y + 4) = -5(x + 1)
y + 4 = -5x - 5
5x + y + 4 + 5 = 0
5x + y + 9 = 0
Equation of normal :
(y - y_{1}) = (-1/m)(x - x_{1})
(y + 4) = (-1/-5)(x - 1)
(y + 4) = (1/5)(x - 1)
5(y + 4) = 1(x - 1)
5y + 20 = x - 1
x - 5y - 20 - 1 = 0
x - 5y - 21 = 0
Problem 6 :
y = 3x^{2} - 1/x at x = -1.
Solution :
y = 3x^{2} - 1/x at x = -1.
When, x = -1
y = 3(-1)^{2} - 1/(-1) = 3 + 1 = 4 |
dy/dx = 6x + 1/x^{2} = 6(-1) + 1/(-1)2 = -6 + 1 = -5 |
So, the required point is (-1, 4).
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - 4) = -5(x + 1)
y - 4 = -5x - 5
5x + y - 4 + 5 = 0
5x + y + 1 = 0
Equation of normal :
(y - y_{1}) = (-1/m) (x - x_{1})
(y - 4) = (-1/-5)(x + 1)
-5(y - 4) = -1(x + 1)
-5y + 20 = -x - 1
x - 5y + 20 + 1 = 0
x - 5y + 21 = 0
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