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Science > Mathematics > Applications of Derivatives > You are Here 
Example – 01:
 Find the equation of the tangent and normal to the curve y = 3x^{2} – x + 1 at point P(1, 3)
 Solution:
Equation of curve is y = 3x^{2} – x + 1 ………… (1)
Differentiating equation (1) w.r.t. x
= 6x – 1
∴ (dy/dx)_{at P(1, 3)} = 6(1) – 1 = 6 – 1 = 5
The slope of the tangent at P(1, 3) is 5 and that of normal is – 1/5.
 Equation of tangent:
Its slope = m = 5
It passes through P(1, 3) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 3 = 5(x – 1)
∴ y – 3 = 5x – 5
∴ 5x – y – 2 = 0
 Equation of normal:
Its slope = m = – 1/5
It passes through P(1, 3) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 3 = ( 1/5)(x – 1)
∴ 5y – 15 = x + 1
∴ x + 5y 15 – 1 = 0
∴ x + 5y – 16 = 0
Ans: Equation of tangent is 5x – y – 2 = 0 and that of normal is x + 5y – 16 = 0
Example – 02:
 Find the equation of the tangent and normal to the curve y = x^{2} + 4x + 1 = 0 at point P(1, 2)
 Solution:
Equation of curve is y = x^{2} + 4x + 1 ………… (1)
Differentiating equation (1) w.r.t. x
= 2x + 4
∴ (dy/dx)_{at P(1, 2)} = 2(1) + 4 = 2 + 4 = 2
The slope of the tangent at P(1, 2) is 2 and that of normal is – 1/2.
 Equation of tangent:
Its slope = m = 2
It passes through P(1, 2) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y + 2 = 2(x + 1)
∴ y + 2 = 2x + 2
∴ 2x – y = 0
 Equation of normal:
Its slope = m = 1/2
It passes through P(1, 2) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y + 2 = ( 1/2)(x + 1)
∴ 2y + 4 = x – 1
∴ x + 2y + 5= 0
Ans: Equation of tangent is 2x – y = 0 and that of normal is X + 2y + 5= 0
Example – 03:
 Find the equation of the tangent and normal to the curve 2x^{2} + 3y^{2} – 5 = 0 at point P(1, 1)
 Solution:
Equation of curve is 2x^{2} + 3y^{2} – 5 = 0 ………… (1)
Differentiating equation (1) w.r.t. x
The slope of the tangent at P(1, 1) is – 2/3 and that of normal is 3/2.
 Equation of tangent:
Its slope = m = – 2/3
It passes through P(1, 1) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 1 =( 2/3)(x – 1)
∴ 3y – 3 = 2x + 2
∴ 2x + 3y – 5 = 0
 Equation of normal:
Its slope = m = 3/2
It passes through P(1, 1) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 1 =(3/2)(x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Ans: Equation of tangent is 2x + 3y – 5 = 0 and that of normal is 3x – 2y – 1 = 0
Example – 04:
 Find the equation of the tangent and normal to the curve x^{2} + y^{3} + xy = 3 at point P(1, 1)
 Solution:
Equation of curve is x^{2} + y^{3} + xy = 3 ………… (1)
Differentiating equation (1) w.r.t. x
The slope of the tangent at P(1, 1) is – 3/4 and that of normal is 4/3.
 Equation of tangent:
Its slope = m = – 4/3
It passes through P(1, 1) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 1 =( 3/4)(x – 1)
∴ 4y – 4 = 3x + 3
∴ 3x + 4y – 7 = 0
 Equation of normal:
Its slope = m = 3/4
It passes through P(1, 1) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 1 =(4/3)(x – 1)
∴ 3y – 3 = 4x – 4
∴ 4x – 3y – 1 = 0
Ans: Equation of tangent is 3x + 4y – 7 = 0 and that of normal is 4x – 3y – 1 = 0
Example – 05:
 Find the equation of the tangent and normal to the curve xy = c^{2} at point P(ct, c/t). Where t is a parameter
 Solution:
Equation of curve is xy = c^{2} ………… (1)
Differentiating equation (1) w.r.t. x
The slope of the tangent at P(ct, c/t) is – 1/t^{2} and that of normal is t^{2}.
 Equation of tangent:
Its slope = m = – 1/t^{2}
It passes through P(ct, c/t) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – c/t =( 1/t^{2})(x – ct)
∴ t^{2} (y – c/t) = – x + ct
∴ y t^{2} – ct = – x + ct
∴ x + y t^{2} – 2ct = 0
 Equation of normal:
Its slope = m = t^{2}
It passes through P(ct, c/t) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – c/t = t^{2}(x – ct)
∴ yt – c = t^{3}(x – ct)
∴ yt – c = x t^{3} – c t^{4}
∴ t^{3}x – yt + c – c t^{4} = 0
Ans: Equation of tangent is x + y t^{2} – 2ct = 0 and that of normal is t^{3}x – yt + c – c t^{4} = 0
Example – 06:
 Find the equation of the tangent and normal to the curve √x – √y = 1 at P(9, 4)
 Solution:
Equation of curve is √x – √y = 1 ………… (1)
Differentiating equation (1) w.r.t. x
The slope of the tangent at P(9, 4) is 2/3 and that of normal is 3/2.
 Equation of tangent:
Its slope = m = 2/3
It passes through P(9, 4) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 4 =(2/3)(x – 9)
∴ 3y – 12 = 2x – 18
∴ 2x – 3y – 6 = 0
 Equation of normal:
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 4 =( 3/2)(x – 9)
∴ 2y – 8 = – 3x + 27
∴ 3x + 2y – 35 = 0
Ans: Equation of tangent is 2x – 3y – 6 = 0 and that of normal is 3x + 2y – 35 = 0
Example – 07:
 Find the equation of the tangent and normal to the curve y = x^{3} – x^{2} – 1 at point whose abscissa is 2.
 Solution:
Equation of curve is y = x^{3} – x^{2} – 1 ………… (1)
Let P be the point whose abscissa (xcoordinate) is 2
Substituting in equation (1)
y = x^{3} – x^{2} – 1 = (2)^{3} – (2)^{2} – 1 = – 8 – 4 – 1 = 13
Hence coordinates of point P are (2, 13)
Differentiating equation (1) w.r.t. x
= 3x^{2} – 2x
∴ (dy/dx)_{at P(2, 13)} = 3(2)^{2} – 2(2) = 12 + 4 = 16
The slope of the tangent at P(2, 13) is 16 and that of normal is – 1/16.
 Equation of tangent:
Its slope = m = 16
It passes through P(2, 13) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y + 13 = 16 (x + 2)
∴ y + 13 = 16x + 32
∴ 16x – y +32 13 = 0
∴ 16x – y + 19 = 0
 Equation of normal:
Its slope = m = 1 1/6
It passes through P(2, 13) ≡ P(x_{1}, y_{1})
slope point form
y – y_{1} = m(x – x_{1})
∴ y + 13 = (1/16) (x + 2)
∴ 16y + 208 = – x – 2
∴ x + 16y + 208 + 2 = 0
∴ x + 16y + 210 = 0
Ans: Equation of tangent is 16x – y + 19 = 0 and that of normal is X + 16y + 210 = 0
Example – 08:
 Find the equation of the tangent and normal to the curve y = x^{2} + 4x at a point whose ordinate is 3.
 Solution:
Equation of curve is y = x^{2} + 4x ………… (1)
Let P be the point whose ordinate (ycoordinate) is 3
Substituting in equation (1)
y = x^{2} + 4x = – 3
∴ x^{2} + 4x + 3 = 0
∴ (x + 3)(x + 1) = 0
∴ x + 3 = 0 or/and x + 1 = 0
∴ x = 3 or/and x = 1
Hence the points are P(3, 3) and Q (1, 3)
Differentiating equation (1) w.r.t. x
= 2x + 4

Let us consider point P(3, 3)
(dy/dx)_{at P(3, 3)} = 2(3) + 4 = – 6 + 4 = – 2
The slopee othe f tangent at P(3, 3) is 2 and that of normal is 1/2.
 Equation of tangent:
Its slope = m = 2
It passes through P(3, 3) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y + 3 = 2 (x + 3)
∴ y + 3 = – 2x – 6
∴ 2x + y + 9 = 0
 Equation of normal:
Its slope = m = 1/2
It passes through P(3, 3) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y + 3 = (1/2)(x + 3)
∴ 2y + 6 = x + 3
∴ x – 2y – 3= 0

Let us consider point P(1, 3)
(dy/dx)_{at P(1, 3)} = 2(2) + 4
The slope of the tangent at P(1, 3) is 2 and that of normal is – 1/2.
 Equation of tangent:
Its slope = m = 2
It passes through P(1, 3) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y + 3 = 2 (x + 1)
∴ y + 3 = 2x + 2
∴ 2x – y – 1 = 0
 Equation of normal:
Its slope = m = – 1/2
It passes through P(1, 3) ≡ P(x_{1}, y_{1})
By slope point form
∴ y – y_{1} = m(x – x_{1})
∴ y + 3 = (1/2)(x + 1)
∴ 2y + 6 = – x – 1
∴ x + 2y + 7 = 0
Ans: At point (3, 3) equation of tangent is 2x + y + 9 = 0 and that of normal is x – 2y – 3= 0
At point (1, 3) equation of tangent is 2x – y – 1 = 0 and that of normal is x + 2y + 7 = 0
Example – 09:
 Find the equation of the tangent and normal to the curve y = x^{2} – 5x at a point where the curve meets the xaxis.
 Solution:
The equation of the curve is y = x^{2} – 5x …………….. (1)
Let P be the point at which the curve meets xaxis (y = 0).
Substituting y = 0 in equation (1)
0 = x^{2} – 5x
∴ x(x – 5) = 0
∴ x = 0 or x = 5
Hence the curve cuts xaxis at P(0, 0) and Q(5, 0)
Differentiating equation (1) w.r.t. x
= 2x – 5

Consider point P(0, 0)
(dy/dx)_{at P(0, 0)} = 2(0) – 5 = 5
The slope of the tangent at P(0, 0) is – 5 and that of normal is 1/5.
 Equation of tangent:
Its slope = m = – 5
It passes through P(0, 0) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 0 = 5 (x – 0)
∴ y = – 5x
∴ 5x + y = 0
 Equation of normal:
Its slope = m = 1/5
It passes through P(0, 0) ≡ P(x_{1}, y_{1})
By slope point form
∴ y – y_{1} = m(x – x_{1})
∴ y – 0 = (1/5) (x – 0)
∴ 5y = x
∴ x – 5y = 0

Consider point P(5, 0)
(dy/dx)_{at P(0, 0)} = 2(5) – 5 = 5
The slope of the tangent at P(0, 0) is 5 and that of normal is – 1/5.
 Equation of tangent:
Its slope = m = 5
It passes through P(5, 0) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 0 = 5 (x – 5)
∴ y = 5x – 25
∴ 5x – y – 25 = 0
 Equation of normal:
Its slope = m = – 1/5
It passes through P(5, 0) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 0 = (1/5) (x – 5)
∴ 5y = – x + 5
∴ x + 5y – 5 = 0
Ans: At point (0, 0) equation of tangent is 5x + y = 0 and that of normal is x – 5y = 0
At point (5, 0) equation of tangent is 5x – y – 25 = 0 and that of normal is x + 5y – 5 = 0
Example – 10:
 Find the equation of the tangent and normal to the curve y = x^{2} + 4x at the point where it cuts the yaxis
 Solution:
Equation of curve is y = x^{2} + 4x ………… (1)
Let P be the point at which the curve cuts yaxis (x = 0).
Substituting x = 0 in equation (1)
y = 0^{2} – 5(0) = 0
Hence the curve cuts xaxis at P(0, 0)
Differentiating equation (1) w.r.t. x
= 2x + 4
(dy/dx)_{at P(0, 0)} = 2(0) + 4 = 0 + 4 = 4
The slope of the tangent at P(0, 0) is 4 and that of normal is – 1/4.
 Equation of tangent:
Its Slope = m = 4
It passes through P(0, 0) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 0 = 4(x – 0)
∴ y = 4x
∴ 4x – y = 0
 Equation of normal:
Its Slope = m = 1/4
It passes through P(0, 0) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 0 = (1/4) (x – 0)
∴ 4y = – x
∴ x + 4y = 0
Example – 11:
 Find the equation of the tangent and normal to the curve y = √2 sin (2x + π/4) at x = π/4
 Solution:
Equation of curve is y = √2 sin (2x + π/4) ………… (1)
Substituting x = π/4 in equation (1)
y = √2 sin (2(π/4) + π/4) = √2 sin (π/2 + π/4) =√2 cos π/4 = √2 x (1/√2 )= 1
Therefore the coordinates of point P are (π/4, 1)
Differentiating equation (1) w.r.t. x
= √2 cos (2x + π/4). 2
∴ = 2√2 cos (2x + π/4)
∴ (dy/dx)_{at P(π/4, 1)} = 2√2 cos (2(π/4) + π/4)
∴ (dy/dx)_{at P(π/4, 1)} = 2√2 cos (π/2 + π/4)
∴ (dy/dx)_{at P(π/4, 1)} = – 2√2 sin (π/4)
∴ (dy/dx)_{at P(π/4, 1)} = – 2√2 x (1/√2 ) = – 2
The slope of the tangent at P(9, 4) is 2 and that of normal is 1/2.
 Equation of tangent:
Its slope = m = 2
It passes through P(π/4, 1) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 1 = 2(x – π/4)
∴ y – 1 = 2x + π/2
∴ 2x + Y – 1 – π/2 = 0
 Equation of normal:
Its slope = m = 1/2
It passes through P(π/4, 1) ≡ P(x_{1}, y_{1})
By slope point form
y – y_{1} = m(x – x_{1})
∴ y – 1 = (1/2)(x – π/4)
∴ 2y – 2 = x – π/4
∴ x – 2y + 2 – π/4 = 0
Ans: Equation of tangent is 2x + Y – 1 – π/2 = 0 and that of normal is x – 2y + 2 – π/4 = 0
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