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Tangents and Normal Tangents and Normal is the introducing part in the Application of Derivatives. The chapter starts with basic concepts of equations of tangent and normal to general curves, angle of intersection between two curves and goes on to discuss more fundamental concepts. The concepts have been explained in detail along with various illustrations and figures (wherever necessary). "Tangents and Normal" is one of the scoring topics of Differential Calculus in the Mathematics syllabus of IIT JEE, AIEEE and other engineering examinations. It requires a good understanding of graphs and further helps in the portion of Coordinate Geometry, another important unit in the syllabus. The tangent to the curve y = f(x) at the point (x, y) makes an angle Ψ with the positive x-axis. Then dy/dx = tan Ψ. Thus the derivative dy/dx or f'(x) represents the slope of the tangent to the curve at the point (x, y). Equation of normal at (x_{1}, y_{1}) where the slope is calculated at the point (x_{1}, y_{1}) In some cases, dy/dx fails to exist but even then a tangent can be drawn. Let us discuss these concepts with the help of an example: PT is the tangent to the curve y = f(x) at the point P(x_{1}, y_{1}). PN is the normal to the curve at P. The slope of the tangent at P(x_{1}, y_{1}) is, [dy/dx]_{(x1,y1)}. The slope of the normal at P(x_{1}, y_{1}) is, -1/[dy/dx]_{(x1,y1)}. Hence the equation of the tangent PT is, y - y_{1} = [dy/dx]_{(x1,y1)} (x - x_{1}), and the equation of the normal PN is, y - y_{1} = -1/[dy/dx]_{(x1,y1)} (x - x_{1}) If (dy/dx)_{(x1,y1)} = 0, then the equation of the normal would be x = x_{1}. If the equation of the curve is in the parametric form x = f(t) and y = g(t), Then dy/dx = dy/dt/dx/dt = (g' (t))/(f' (t)). Thus the equations of the tangent and the normal are y - g(t) = (g' (t))/(f' (t)) (x - f(t)) and f'(t)[x - f(t)] + g'(t)[y - g(t)] = 0. If dy/dx at (x_{1}, y_{1}) = 0 then the tangent is parallel to x-axis and conversely. If the tangent is parallel to ax + by + c = 0 then dy/dx = -a/b If dy/dx at (x_{1}, y_{1}) → ∞ i.e. dx/dy at (x_{1}, y_{1}) = 0 then the tangent is perpendicular to x- axis. If the tangent at P(x_{1}, y_{1}) is equally inclined to the coordinate axes, then dy/dx at (x_{1}, y_{1}) = ± 1. If the tangent makes equal non-zero intercepts on the coordinate axis, then dy/dx at (x_{1}, y_{1}) = -1. If the tangent cuts off from the coordinate axis equal distance from the origin then dy/dx = ±1. For x^{2/3} + y^{2/3} = a^{2/3}, assume the parametric coordinates x = a cos^{3}θ and y = a sin^{3}θ . For √x + √y = √a, assume x = cos^{4}θ and y = a sin^{4}θ For x^{n}/a^{n} + y^{n}/b^{n} = 1, assume x = a (sin θ)^{2/n} and y = b(sin θ)^{2/n}. For y^{2} = x^{3}, take x = t^{2} and y = t^{3}. The angle of intersection of two curves at a point P is defined as the angle between the two tangents to the curve at their point of intersection. Let there be two curves y = f_{1}(x) and y = f_{2}(x) which intersect each other at point (x_{1}, y_{1}). If we draw tangents to these curves at the intersecting point, the angle between these tangents is called the angle between two curves. Let m_{1} = (df_{1}(x))/dx |_{(x = x1)} and m_{2} = (df_{2}(x))/dx |_{(x = x1)} And both m_{1} and m_{2} are finite. The acute angle between the curves is given by θ = tan^{-1} |(m_{1} – m_{2})/(1 + m_{1} m_{2})| Note: (p - q) is also an angle between lines. If m_{1} (or m_{2}) is infinity the angle is given by θ = |π/2 – θ_{1}| where θ_{1} = tan^{-1} m_{2} (or tan^{-1} m_{1}) In the figure given below, Φ is the angle between the two curves,which is given by Φ = Ψ_{1} - Ψ_{2} ⇒ tan Φ = tan (Ψ_{1} - Ψ_{2}) = (tan Ψ_{1} – tan Ψ_{2})/(1 + tan Ψ_{1} tan Ψ_{2} ), where tan Ψ_{1} = f'(x_{1}) and tan Ψ_{2} = g'(x_{1}). Two curves are said to cut each other orthogonally if the angle between them is a right angle, in which case we will have, tanΨ_{1} tanΨ_{2} = -1. Two curves touch each other if the angle between the tangents to the curves at the point of intersection is 0^{o}, in which case we will have, tanΨ_{1} = tanΨ_{2}. Illustration: Show that the curves ax^{2} + by^{2} = 1 and cx^{2} + dy^{2} = 1 cut each other orthogonally if, 1/a – 1/b = 1/c - 1/d. Solution: Let the two curves cut each other at the point (x_{1}, y_{1}). Then ax_{1}^{2} + by_{1}^{2} = 1 ...... (1) and cx_{1}^{2} + dy_{1}^{2} = 1. ...... (2) From (1) and (2), we get (a - c)x_{1}^{2} + (b - d)y_{1}^{2} = 0. ...... (3) Slope of the tangent to the curve ax^{2} + by^{2} = 1, at (x_{1}, y_{1}) is given by tan Ψ_{1} = [dy/dx]_{(x1,y1)} = -ax_{1}/by_{1} . Slope of the tangent to the curve cx^{2} + dy^{2} = 1 at (x_{1}, y_{1}) is given by tan Ψ_{2} = [dy/dx]_{(x1,y1)} = -cx_{1}/dy_{1}. If the two curves cut orthogonally, we must have, (-ax_{1}/by_{1})(-cx_{1}/dy_{1}) = -1 ⇒ acx_{1}^{2} + bdy_{1}^{2} = 0. ...... (4) From (3) and (4) we have (a-c)/ac = (b-d)/bd ⇒ 1/a – 1/c = 1/c – 1/d. Illustration 2: Prove that the tangent lines to the curve y^{2} = 4ax at points where x = a are at right angles to each other. Solution 2: y^{2} = 4ax When x = a y^{2} = 4a^{2} ⇒ y = ± 2a Points are (a, 2a) and (a, - 2a) Slope of the tangent of the curve y^{2} = 4ax is y dy/dx = 4a dy/dx = 2a/y for y = 2a, dy/dx = 1 = m_{1} for y = -2a dy/dx = - 1 = m_{2} m_{1}m_{2} = -1 Let P(x, y) be any point on y = f(x). Let the tangent drawn at 'P' meet the x-axis at 'T', and normal drawn at 'P' meets the x-axis at 'N'. PT is called the length of the tangent and PN is called the length of the normal. If 'P_{1}' be the projection of the point P on the x-axis then TP_{1} is called the sub-tangent (projection of line segment PT on the x-axis) and NP_{1} is called the sub normal (projection of line segment PN on the x-axis). Tangent: TP = PP_{1} cosec θ = y √(1 + cot^{2}θ) = Subtangent: TP_{1} = PP_{1} cot θ = Normal: PN = PP_{1} sec θ = y √(1 + tan^{2}θ) = Subnormal: NP_{1} = PP_{1} tan y = |y(dy/dx)| Interpretation of dy/dt as a rate measure: We know that by the derivative dr/dt, we mean the rate of change of distance r with respect to time t. In the similar way, when some quantity y varies or changes with some other quantity x while satisfying some rule y = f(x), then the derivative dy/dx i.e. f’(x) implies the rate of change of y with respect to x. Let us discuss some of the solved examples based on these concepts: Example 1: The coordinates of the feet of normals drawn from the point (14,7) to the curve y^{2 }- 16x – 8y = 0 are (a) (0,0) (b) 3,2) (c ) (3,-4) (d) (8, 16) Solution: Given curve is y^{2 }- 16x – 8y = 0 ….… (1) Let P = (14, 7) Then equation (1) can be written as y^{2} – 8y = 16x or y^{2} – 8y + 16 = 16x + 16 Or (y – 4)^{2} = 16(x + 1) ….….…. (2) This is of the form (y –l)^{2} = 4a(x-s), where a = 4, s = -1, l = 4. Let (-1+4t^{2}, 4+8t) be any point on the curve (2). Then from (2) , we have 2(y-4) dy/dx = 16. Hence, dy/dx = 8/(y-4) at (4t^{2} – 1, 4 + 8t), dy/dx = 1/t Hence, the equation of normal at (-1 + 4t^{2}, 4+8t) is y - 4 – 8t = – t(x + 1 – 4t^{2}) or tx + y – 4 – 8t + t – 4t^{3} = 0 ….….. (3) If line passes through the point P(14, 7) then 14t + 7 – 4t^{3} – 7t – 4 = 0 or 4t^{3} – 7t -3 = 0 (t + 1)(4t^{2} – 4t – 3) = 0 Hence, t = -1, (4 ± 8)/8 = -1, 3/2 , -1/2. when t = -1, foot of the normal is (3, -4) when t = 3/2, foot of the normal is (8, 16) when t = -1/2, foot of the normal is (0, 0). Example 2: Find the equations of the tangents to the curve x^{2} + y^{2} – 2x – 4y + 1 = 0 which are parallel to the x-axis. Solution: The equation of the curve is x^{2} + y^{2} – 2x – 4y + 1 = 0. Hence, 2x + 2y dy/dx – 2 – 4 dy/dx = 0 So, (2y – 4) dy/dx = 2 – 2x Hence, dy/dx = (2 – 2x)/(2y - 4) = (1 – x)/(y – 2) Since the tangents are parallel to the x- axis, so the slope of each of the tangent is zero. Hence, (1 - x)/(y – 2) = 0 So, x = 1. At x = 1, 1^{2} + y^{2} – 2(1) – 4y + 1 = 0 So, y^{2} – 4y = 0 this gives y = 0 or y = 4. Therefore, the points are (1,0) and (1,4). Hence, the equation of tangent through (1, 0) and parallel to the x-axis is y = 0 and the equation of tangent through (1, 4) and parallel to the x-axis is y = 4. Hence, the required equations of the tangents are y = 0 and y = 4. Q1. If the curves are orthogonal, then (a) (dy_{1}/dx)(dy_{2}/dx) = -1 (b) (dy_{1}/dx)(dy_{2}/dx) = 1 (c) (dy_{1}/dx)(dy_{2}/dx) = 0 (d) (dy_{1}/dx)^{2} = -1 Q2. If θ is the angle between y = x^{2} and 6y = 7-x^{3} at (a, a) then θ is: (a) 2π (b) π (c) π/3 (d) π/2 Q3. If F(x) = f(x).g(x) are such that f(x) is continuous at x = a and g(x) is differentiable at x = a with g(a) = 0 then the product function f(x).g(x) is (a) continuous at x = a (b) differentiable at x = a (c) can’t say (d) may or may not be differentiable at x = a Q4. If the curves ay + x^{2 }= 7, a > 0 and x^{3} = y cut orthogonally at (1,1), then a = ? (a) 4 (b) 1/2 (c) 6 (d) 2 Q5. If the value of dy/dx at the point (x_{1},y_{1}) and exists and is 0, then (a) tangent is parallel to x-axis. (b) tangent is perpendicular to x-axis. (c) tangent may or may not be parallel to x-axis. (d) can’t say

Tangents and Normal is the introducing part in the Application of Derivatives. The chapter starts with basic concepts of equations of tangent and normal to general curves, angle of intersection between two curves and goes on to discuss more fundamental concepts. The concepts have been explained in detail along with various illustrations and figures (wherever necessary).

"Tangents and Normal" is one of the scoring topics of Differential Calculus in the Mathematics syllabus of IIT JEE, AIEEE and other engineering examinations. It requires a good understanding of graphs and further helps in the portion of Coordinate Geometry, another important unit in the syllabus.

The tangent to the curve y = f(x) at the point (x, y) makes an angle Ψ with the positive x-axis. Then dy/dx = tan Ψ. Thus the derivative dy/dx or f'(x) represents the slope of the tangent to the curve at the point (x, y).

Equation of normal at (x_{1}, y_{1})

where the slope is calculated at the point (x_{1}, y_{1})

In some cases, dy/dx fails to exist but even then a tangent can be drawn.

Let us discuss these concepts with the help of an example:

PT is the tangent to the curve y = f(x) at the point P(x_{1}, y_{1}).

PN is the normal to the curve at P.

The slope of the tangent at P(x_{1}, y_{1}) is, [dy/dx]_{(x1,y1)}.

The slope of the normal at P(x_{1}, y_{1}) is, -1/[dy/dx]_{(x1,y1)}.

Hence the equation of the tangent PT is, y - y_{1} = [dy/dx]_{(x1,y1)} (x - x_{1}),

and the equation of the normal PN is, y - y_{1} = -1/[dy/dx]_{(x1,y1)} (x - x_{1})

If (dy/dx)_{(x1,y1)} = 0, then the equation of the normal would be x = x_{1}.

If the equation of the curve is in the parametric form x = f(t) and y = g(t),

Then dy/dx = dy/dt/dx/dt = (g' (t))/(f' (t)).

Thus the equations of the tangent and the normal are

y - g(t) = (g' (t))/(f' (t)) (x - f(t))

and f'(t)[x - f(t)] + g'(t)[y - g(t)] = 0.

If dy/dx at (x_{1}, y_{1}) = 0 then the tangent is parallel to x-axis and conversely.

If the tangent is parallel to ax + by + c = 0 then dy/dx = -a/b

If dy/dx at (x_{1}, y_{1}) → ∞ i.e. dx/dy at (x_{1}, y_{1}) = 0 then the tangent is perpendicular to x- axis.

If the tangent at P(x_{1}, y_{1}) is equally inclined to the coordinate axes, then dy/dx at (x_{1}, y_{1}) = ± 1.

If the tangent makes equal non-zero intercepts on the coordinate axis, then dy/dx at (x_{1}, y_{1}) = -1.

If the tangent cuts off from the coordinate axis equal distance from the origin then dy/dx = ±1.

For x^{2/3} + y^{2/3} = a^{2/3}, assume the parametric coordinates x = a cos^{3}θ and y = a sin^{3}θ .

For √x + √y = √a, assume x = cos^{4}θ and y = a sin^{4}θ

For x^{n}/a^{n} + y^{n}/b^{n} = 1, assume x = a (sin θ)^{2/n} and y = b(sin θ)^{2/n}.

For y^{2} = x^{3}, take x = t^{2} and y = t^{3}.

The angle of intersection of two curves at a point P is defined as the angle between the two tangents to the curve at their point of intersection.

Let there be two curves y = f_{1}(x) and y = f_{2}(x) which intersect each other at point (x_{1}, y_{1}). If we draw tangents to these curves at the intersecting point, the angle between these tangents is called the angle between two curves.

Let m_{1} = (df_{1}(x))/dx |_{(x = x1)} and m_{2} = (df_{2}(x))/dx |_{(x = x1)}

And both m_{1} and m_{2} are finite.

The acute angle between the curves is given by

θ = tan^{-1} |(m_{1} – m_{2})/(1 + m_{1} m_{2})|

Note: (p - q) is also an angle between lines.

If m_{1} (or m_{2}) is infinity the angle is given by θ = |π/2 – θ_{1}| where

θ_{1} = tan^{-1} m_{2} (or tan^{-1} m_{1})

In the figure given below, Φ is the angle between the two curves,which is given by

Φ = Ψ_{1} - Ψ_{2}

⇒ tan Φ = tan (Ψ_{1} - Ψ_{2})

= (tan Ψ_{1} – tan Ψ_{2})/(1 + tan Ψ_{1} tan Ψ_{2} ), where tan Ψ_{1} = f'(x_{1}) and tan Ψ_{2} = g'(x_{1}).

Two curves are said to cut each other orthogonally if the angle between them is a right angle, in which case we will have,

tanΨ_{1} tanΨ_{2} = -1.

Two curves touch each other if the angle between the tangents to the curves at the point of intersection is 0^{o}, in which case we will have,

tanΨ_{1} = tanΨ_{2}.

Illustration:

Show that the curves ax^{2} + by^{2} = 1 and cx^{2} + dy^{2} = 1 cut each other orthogonally if, 1/a – 1/b = 1/c - 1/d.

Solution:

Let the two curves cut each other at the point (x_{1}, y_{1}). Then

ax_{1}^{2} + by_{1}^{2} = 1 ...... (1)

and cx_{1}^{2} + dy_{1}^{2} = 1. ...... (2)

From (1) and (2), we get

(a - c)x_{1}^{2} + (b - d)y_{1}^{2} = 0. ...... (3)

Slope of the tangent to the curve ax^{2} + by^{2} = 1, at (x_{1}, y_{1}) is given by

tan Ψ_{1} = [dy/dx]_{(x1,y1)} = -ax_{1}/by_{1} .

Slope of the tangent to the curve cx^{2} + dy^{2} = 1 at (x_{1}, y_{1}) is given by

tan Ψ_{2} = [dy/dx]_{(x1,y1)} = -cx_{1}/dy_{1}. If the two curves cut orthogonally, we must have,

(-ax_{1}/by_{1})(-cx_{1}/dy_{1}) = -1 ⇒ acx_{1}^{2} + bdy_{1}^{2} = 0. ...... (4)

From (3) and (4) we have

(a-c)/ac = (b-d)/bd ⇒ 1/a – 1/c = 1/c – 1/d.

Illustration 2:

Prove that the tangent lines to the curve y^{2} = 4ax at points where x = a are at right angles to each other.

Solution 2:

y^{2} = 4ax

When x = a

y^{2} = 4a^{2}

⇒ y = ± 2a

Points are (a, 2a) and (a, - 2a)

Slope of the tangent of the curve y^{2} = 4ax is

y dy/dx = 4a

dy/dx = 2a/y

for y = 2a, dy/dx = 1 = m_{1}

for y = -2a dy/dx = - 1 = m_{2}

m_{1}m_{2} = -1

Let P(x, y) be any point on y = f(x). Let the tangent drawn at 'P' meet the x-axis at 'T', and normal drawn at 'P' meets the x-axis at 'N'. PT is called the length of the tangent and PN is called the length of the normal.

If 'P_{1}' be the projection of the point P on the x-axis then TP_{1} is called the sub-tangent (projection of line segment PT on the x-axis) and NP_{1} is called the sub normal (projection of line segment PN on the x-axis).

Tangent:

TP = PP_{1} cosec θ = y √(1 + cot^{2}θ)

=

Subtangent:

TP_{1} = PP_{1} cot θ

Normal:

PN = PP_{1} sec θ = y √(1 + tan^{2}θ)

Subnormal:

NP_{1} = PP_{1} tan y = |y(dy/dx)|

We know that by the derivative dr/dt, we mean the rate of change of distance r with respect to time t. In the similar way, when some quantity y varies or changes with some other quantity x while satisfying some rule y = f(x), then the derivative dy/dx i.e. f’(x) implies the rate of change of y with respect to x.

Let us discuss some of the solved examples based on these concepts:

Example 1: The coordinates of the feet of normals drawn from the point (14,7) to the curve y^{2 }- 16x – 8y = 0 are

(a) (0,0) (b) 3,2) (c ) (3,-4) (d) (8, 16)

Solution: Given curve is y^{2 }- 16x – 8y = 0 ….… (1)

Let P = (14, 7)

Then equation (1) can be written as y^{2} – 8y = 16x

or y^{2} – 8y + 16 = 16x + 16

Or (y – 4)^{2} = 16(x + 1) ….….…. (2)

This is of the form (y –l)^{2} = 4a(x-s), where a = 4, s = -1, l = 4.

Let (-1+4t^{2}, 4+8t) be any point on the curve (2).

Then from (2) , we have 2(y-4) dy/dx = 16.

Hence, dy/dx = 8/(y-4)

at (4t^{2} – 1, 4 + 8t), dy/dx = 1/t

Hence, the equation of normal at (-1 + 4t^{2}, 4+8t) is

y - 4 – 8t = – t(x + 1 – 4t^{2})

or tx + y – 4 – 8t + t – 4t^{3} = 0 ….….. (3)

If line passes through the point P(14, 7) then

14t + 7 – 4t^{3} – 7t – 4 = 0

or 4t^{3} – 7t -3 = 0

(t + 1)(4t^{2} – 4t – 3) = 0

Hence, t = -1, (4 ± 8)/8 = -1, 3/2 , -1/2.

when t = -1, foot of the normal is (3, -4)

when t = 3/2, foot of the normal is (8, 16)

when t = -1/2, foot of the normal is (0, 0).

Example 2: Find the equations of the tangents to the curve x^{2} + y^{2} – 2x – 4y + 1 = 0 which are parallel to the x-axis.

Solution: The equation of the curve is x^{2} + y^{2} – 2x – 4y + 1 = 0.

Hence, 2x + 2y dy/dx – 2 – 4 dy/dx = 0

So, (2y – 4) dy/dx = 2 – 2x

Hence, dy/dx = (2 – 2x)/(2y - 4) = (1 – x)/(y – 2)

Since the tangents are parallel to the x- axis, so the slope of each of the tangent is zero.

Hence, (1 - x)/(y – 2) = 0

So, x = 1.

At x = 1, 1^{2} + y^{2} – 2(1) – 4y + 1 = 0

So, y^{2} – 4y = 0

this gives y = 0 or y = 4.

Therefore, the points are (1,0) and (1,4).

Hence, the equation of tangent through (1, 0) and parallel to the x-axis is y = 0 and the equation of tangent through (1, 4) and parallel to the x-axis is y = 4.

Hence, the required equations of the tangents are y = 0 and y = 4.

Q1. If the curves are orthogonal, then

(a) (dy_{1}/dx)(dy_{2}/dx) = -1

(b) (dy_{1}/dx)(dy_{2}/dx) = 1

(c) (dy_{1}/dx)(dy_{2}/dx) = 0

(d) (dy_{1}/dx)^{2} = -1

Q2. If θ is the angle between y = x^{2} and 6y = 7-x^{3} at (a, a) then θ is:

(a) 2π

(b) π

(c) π/3

(d) π/2

Q3. If F(x) = f(x).g(x) are such that f(x) is continuous at x = a and g(x) is differentiable at x = a with g(a) = 0 then the product function f(x).g(x) is

(a) continuous at x = a

(b) differentiable at x = a

(c) can’t say

(d) may or may not be differentiable at x = a

Q4. If the curves ay + x^{2 }= 7, a > 0 and x^{3} = y cut orthogonally at (1,1), then a = ?

(a) 4

(b) 1/2

(c) 6

(d) 2

Q5. If the value of dy/dx at the point (x_{1},y_{1}) and exists and is 0, then

(a) tangent is parallel to x-axis.

(b) tangent is perpendicular to x-axis.

(c) tangent may or may not be parallel to x-axis.

(d) can’t say

The chapter is important not only because it fetches 2-3 questions in most of the engineering examination but also because it is a helping hand in the unit of Coordinate Geometry.Tangents and Normal is an important chapter in Differential calculus. It is considered to be marks fetching as the Multiple Choice Questions that are framed on this topic are direct and simple. You are expected to do all the questions based on this to take an edge in IIT JEE examination. It is very important to master these concepts as this forms a strong base of your preparation for IIT JEE, AIEEE, DCE, EAMCET and other engineering entrance examinations.

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