Equations of Tangents and Normals – 02

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Example – 12:

• Find the equation of the tangent and normal to the curve x = sin θ and y = cos 2θ at θ = π/6
• Solution:

The equation of the curve is x = sin θ and y = cos 2θ ………. (1)

Parameter θ = π/6

∴  x = sin π/6 = 1/2 and y = cos 2(π/6) = cos π/3 = ½

Therefore the point on the curve is P(1/2, 1/2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1/2, 1/2) is -2 and that of normal is 1/2.

• Equation of tangent:

Its slope = m = -2

It passes through P(1/2, 1/2) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – 1/2 = -2(x – 1/2)

∴  y – 1/2 = -2x + 1

∴  2y – 1 = -4x + 2

∴  4x + 2y – 3 = 0

• Equation of normal:

By slope point form

y – y1 = m(x – x1)

∴  y – 1/2 = (1/2)(x – 1/2)

∴  2y – 1 = x – 1/2

∴  4y – 2 = 2x – 1

∴  2x – 4y + 1= 0

Ans: Equation of tangent is 4x + 2y – 3 = 0 and that of normal is 2x – 4y + 1= 0

Example – 13:

• Find the equation of the tangent and normal to the curve x = acos3θ and y = asin3θ at θ = π/4
• Solution:

Equation of curve is x = acos3θ and y = asin3θ ………. (1)

Parameter θ = π/4

∴  x = acos3 (π/4) = a(1/2)3 = a (1/22) = a/22

∴  y = asin3 (π/4) = a(1/2)3 = a (1/22) = a/22

Therefore the point on the curve is P(a/22, a/22)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(a/22, a/22) is – 1 and that of normal is 1.

• Equation of tangent:

Its slope = m = – 1

It passes through P(a/22, a/22) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – a/22 = -1(x – a/22)

∴  22 y – a = – 22 x + a

∴  22 x + 22 y – 2a = 0

∴  √2 x + 2 y –a = 0

• Equation of normal:

By slope point form

y – y1 = m(x – x1)

∴  y – a/22 = 1(x – a/22)

∴  y – a/22 = x – a/22

∴  22 y – a = 22 x – a

∴  22 x – 22 y = 0

∴  x – y = 0

Ans: Equation of tangent is 2 x + 2 y – a = 0 and that of normal is x – y = 0

Example – 14:

• Find the equation of the tangent and normal to the curve x = a sec θ and y = a tan θ at θ = π/6
• Solution:

Equation of curve is x = a sec θ and y = a tan θ ………. (1)

Parameter θ = π/6

∴  x = a sec (π/6) = 2a/3 and y = a tan (π/6) = a/3

Therefore the point on the curve is P(2a/3, a/3)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(2a/3, a/3) is 2 and that of normal is – 1/2.

• Equation of tangent:

Its slope = m = 2

It passes through P(2a/3, a/3) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – a/3 = 2(x – 2a/3)

∴  y – a/3 = 2x – 4a/3

∴  √3 y – a  = 2 3 x – 4a

∴  23 x – 3 y – 3a = 0

∴  2x – y –  3a = 0

• Equation of normal:

By slope point form

y – y1 = m(x – x1)

∴  y – a/3 = (-1/2)(x – 2a/3)

∴  2y – 2a/3= – x + 2a/3

∴  23 y – 2a  = – 3 x + 2a

∴  √3 x + 23 y – 4a = 0

Ans: Equation of tangent is 2x – y –  3 a = 0 and that of normal is 3 x + 23 y – 4a = 0

Example – 15:

• Find the equation of the tangent to the curve x = a(θ  + sin θ) and y = a(1  + cos θ) at θ = π/2
• Solution:

Equation of curve is x = a(θ  + sin θ) and y = a(1  + cos θ) ………. (1)

Parameter θ = π/6

∴  x = a(θ  + sin θ) = a(π/2 + sin π/2)=a(π/2 + 1) and

∴  y = a(1  + cos θ) = a(1 + cos π/2) = a(1 + 0) = a

Therefore the point on the curve is P(a(π/2 + 1), a)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(a(π/2 + 1), a) is -1 and that of normal is 1.

• Equation of tangent:

Its slope = m = -1

It passes through P(a(π/2 + 1), a) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – a = -1(x – a(π/2 + 1))

∴  y – a = – x + a(π/2) + a

∴  x + y – a – a(π/2) – a = 0

∴  x – y – a(π/2)  – 2a = 0

Ans: Equation of tangent is x – y – a(π/2)  – 2a = 0

Example – 16:

• Find the equation of tangent and normal to the curve x = 1/t and y = t – 1/t at t = 2
• Solution:

Equation of curve is x = 1/t and y = t – 1/t ………. (1)

Parameter t = 2

∴  x = 1/2 and y = 2 – 1/2 = 3/2

Therefore the point on the curve is P(1/2, 3/2)

Differentiating equation (1) w.r.t. x

The slope of the tangent at P(1/2, 3/2) is -5 and that of normal is 1/5.

• Equation of tangent:

Its slope = m = – 5

It passes through P(1/2, 3/2) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – 3/2   = – 5(x – 1/2)

∴  y – 3/2   = – 5x + 5/2

∴  5x + y – 3/2 – 5/2 = 0

∴  5x + y – 4 = 0

• Equation of normal:

By slope point form

y – y1 = m(x – x1)

∴  y – 3/2   = 1/5(x – 1/2)

∴  5y – 15/2 = x – 1/2

∴  x – 5y – 1/2 + 15/2 = 0

∴  x – 5y + 7 = 0

Ans: Equation of tangent is 5x + y – 4 = 0 and that of normal is x – 5y + 7 = 0

Example – 17:

• Show that the tangent to the curve 8y = (x -2)2 at the point (-6, 8) is parallel to the tangent to the curve y = x + 3/x at point (1, 4)
• Solution:

Consider the first curve 8y = (x -2)2

Differentiating both sides w.r.t. x

8 = 2(x – 2) (1)

= (x – 2)/4

∴  (dy/dx)at P(-6, 8) = (- 6 – 2)/4 = -8/4 = -2

Thus slope of tangent to curve 8y = (x -2)2 at the point (-6, 8) is – 2 ………….. (1)

Consider the second curve y = x + 3/x

Differentiating both sides w.r.t. x

= 1 – 3/x2

∴  (dy/dx)at Q(1, 4) =  1- 3/12 = 1- 3 = -2

Thus slope of tangent to curve y = x + 3/x at the point (1, 4) is – 2 ………….. (2)

From statements (1) and (2) we can see that the two slopes are equal.

Hence the two tangents are parallel to each other (proved as required)

Example – 18:

• Show that the tangents to the curve x2 = 2y and 6y = 5 – 2x3 at x = 1 are perpendicular to each other.
• Solution:

Consider the first curve x2 = 2y

Differentiating both sides w.r.t. x

2x = 2

∴  = x

∴  (dy/dx)at x = 1 = 1

Thus slope of tangent to curve x2 = 2y at x = 1 is 1 ………….. (1)

Consider the second curve 6y = 5 – 2x3

Differentiating both sides w.r.t. x

6 = -6x2

∴  = -x2

∴  (dy/dx)at x = 1 =  -(1)2 = -1

Thus slope of tangent to curve 6y = 5 – 2x3 at x = 1 is -1 ………….. (2)

From statements (1) and (2) we can see that the product of the two slopes is -1.

Hence the two tangents are perpendicular to each other (proved as required)

Example – 19:

• Find the points on the curve y = 3x2 – 9x + 8 where the tangent is parallel to x- axis
• Solution:

The tangent is parallel to x-axis hence its slope m = (dy/dx) = 0

The equation of the curve is y = 3x2 – 9x + 8  …………… (1)

Differentiating both sides w.r.t. x

= 6x – 9

Thus 6x – 9 = 0

∴  6x = 9

∴  x = 3/2

Substituting in equation (1)

y = 3(3/2)2 – 9(3/2) + 8= 3(9/4) – 27/2 + 8

∴  y = 27/4 -27/2 + 8 = -27/4 + 8 = 5/4

The point on the curve is (3/2, 5/4)

Example – 20:

• Find the points on the curve y = 3x2 – 9x + 8 where the tangent makes an angle 45o with the x- axis
• Solution:

The tangent makes an angle 45o with the x- axis

Slope of tangent = m = tan 45o = 1 = (dy/dx)

The equation of the curve is y = 3x2 – 9x + 8  …………… (1)

Differentiating both sides w.r.t. x

= 6x – 9

Thus 6x – 9 = 1

∴  6x = 10

∴  x = 5/3

Substituting in equation (1)

y = 3(5/3)2 – 9(5/3) + 8= 3(25/9) – 15 + 8

∴  y = 25/3 – 15 + 8 = 25/3 – 7 = 4/3

The point on the curve is (5/3, 4/3)

Example – 21:

• Find the point on the curve y = x – 3 where the tangent is perpendicular to the line 6x + 3y – 5 = 0
• Solution:

Equation of given line 6x + 3y – 5 = 0

Slope of given line = – (coefficient of x/coefficient of y) = – (6/3) = – 2

As the tangent is perpendicular to given line

Slope of tangent = 1/2

Equation of curve is y = x – 3

Differentiating both sides w.r.t. x

squaring both sides

1 = x – 3

∴  x = 4

Substituting in equation (1)

y = 4 – 3 = 1  = ±1

The points on the curve is (4, 1) and (4, -1)

Example – 22:

• Find the coordinates of the point on the curve y = x – 4/x, where the tangent at that point is parallel to the line y = 2x.
• Solution:

Equation of given line y = 2x

Slope of given line = 2

Now the tangent is parallel to the given line

Slope of tangent = m = 2 = (dy/dx)

Equation of the curve is y = x – 4/x

Differentiating both sides w.r.t.x

= 1 – 4(- 1/x2) = 1 + 4/x2

∴  1 + 4/x2 = 2

∴  4/x2 = 1

∴  x2 = 4

∴  x = ± 2

When x = 2

y = 2 – 4/2 = 2 – 2 = 0

Hence the point is (2, 0)

When x = -2

y = – 2 – 4/(-2) = – 2 + 2 = 0

Hence the point is (- 2, 0)

The required points are (2, 0) and (-2, 0)

Example – 23:

• If the line x + y = 0 touches the curve 2y2 = ax2 + b at (1, -1) find a and b
• Solution:

Equation of tangent is x + y = 0

Slope of tangent = – (coefficient of x/coefficient of y) = – (1/1) = -1

Equation of curve is 2y2 = ax2 + b

Differentiating both sides w.r.t. x

4y = 2ax

∴  =ax/2y

∴  (dy/dx)at P(1, -1) = a(1)/2(-1) = – a/2

– a/2 = -1

∴  a = 2

Point (1, -1) lies on the curve 2y2 = ax2 + b

2(-1)2 = 2(1)2 + b

∴  2 = 2 + b

∴  b = 0

Ans: a = 2 and b = 0

Example – 24:

• For the curve y = 3x – 2x2, find the equation of the tangent whose slope is – 1.
• Solution:

Slope of tangent = -1

Equation of curve is y = 3x – 2x2   ………. (1)

Differentiating both sides w.r.t. x

= 3 – 4x

∴  3 – 4x = -1

∴  4 = 4x

∴  x = 1

Substituting in equation (1)

y = 3(1) – 2(1)2 = 3 – 2 = 1

The tangent touches the curve at (1, 1) ≡ P(x1, y1) and its slope = m = -1

By slope point form

y – y1 = m(x – x1)

∴  y – 1 = – 1(x – 1)

∴  y – 1 = – x + 1

∴ x + y – 2 = 0

Ans: The equation of tangent is x + y – 2 = 0

Example – 25:

• Find equations of tangents and normal to the curve y = 6 – x2, where the normal is parallel to the cline x – 4y + 3 = 0
• Solution:

Equation of given line x – 4y + 3 = 0

Slope of given line = – (coefficient of x/coefficient of y) = – (1/-4) = 1/4

As the normal is perpendicular to given line

Slope of normal = 1/4

Slope of tangent = – 4

The equation of curve isy = 6 – x2

Differentiating both sides w.r.t. x

= – 2x = – 4

x = 2

Substituting in equation of curve

y = 6 – 22 = 6 – 4 = 2

The tangent touches the curve at (2, 2)

• Equation of tangent:

Its slope = m = – 4

It passes through P(2, 2) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – 2 = – 4(x – 2)

∴  y – 2 = – 4x + 8

∴  4x + y – 10 = 0

• Equation of normal:

Its slope = m = 1/4

It passes through P(2, 2) ≡ P(x1, y1)

By slope point form

y – y1 = m(x – x1)

∴  y – 2 = 1/4(x – 2)

∴  4y – 8 = x – 2

∴  x – 4y + 6 = 0

Ans: Equation of tangent is 4x + y – 10 = 0 and that of normal is x – 4y + 6 = 0

Example – 25:

• If the line y = 4x -5 touches the curve y2 = ax3 + b at the point (2, 3) show that 7a + 2b = 0
• Solution:

Equation of tangent is 4x – y -5 = 0

Slope of the tangent = – (coefficient of x/coefficient of y) = – (4/-1) = 4 = (dy/dx)

Equation of curve is

y2 = ax3 + b

Differentiating both sides w.r.t. x

2y(dy/dx) = 3ax2

∴  (dy/dx) = 3ax2/2y

∴  (dy/dx) at(2, 3) = 3a(2)2/ 2(3)= 2a

Thus slope of tangent to the curve at (2, 3) is 2a

2a = 4

∴  a = 2

Now point (2, 3) lies on the curve y2 = ax3 + b

32 = 2(2)3 + b

∴  9 = 16 + b

∴   b = 9 – 16 = -7

To prove that 7a + 2b = 0

L.H.S. = 7a + 2b = 7(2) + 2(-7) = 14 – 14 = 0 = R.H.S.

∴ 7a + 2b = 0 (Proved as required)

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