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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 JEEAIEEE 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).

Tangent to a Curve

Equation of normal at (x1, y1)

y-y_{1} = \frac{-1}{\frac{dy}{dx}} (x-x_{1})   where the slope is calculated at the point (x1, y1)

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(x1, y1).

PN is the normal to the curve at P.

The slope of the tangent at P(x1, y1) is, [dy/dx](x1,y1).

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

Tangent and Normal to a curve

Hence the equation of the tangent PT is, y - y1 = [dy/dx](x1,y1) (x - x1),

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

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

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.

  1. If dy/dx at (x1, y1) = 0 then the tangent is parallel to x-axis and conversely.

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

  3. If dy/dx at (x1, y1) → ∞ i.e. dx/dy at (x1, y1) = 0 then the tangent is perpendicular to x- axis.

  4. If the tangent at P(x1, y1) is equally inclined to the coordinate axes, then dy/dx at (x1, y1) = ± 1.

  5. If the tangent makes equal non-zero intercepts on the coordinate axis, then dy/dx at (x1, y1) = -1.

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

  1. For x2/3 + y2/3 = a2/3, assume the parametric coordinates x = a cos3θ and y = a sin3θ .

  2. For √x + √y = √a, assume x = cos4θ and y = a sin4θ

  3. For xn/an + yn/bn = 1, assume x = a (sin θ)2/n and y = b(sin θ)2/n.

  4. For y2 = x3, take x = t2 and y = t3.

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 = f1(x) and y = f2(x) which intersect each other at point (x1, y1). If we draw tangents to these curves at the intersecting point, the angle between these tangents is called the angle between two curves.

Let m1 = (df1(x))/dx |(x = x1) and m2 = (df2(x))/dx |(x = x1)

And both m1 and m2 are finite.

The acute angle between the curves is given by

θ = tan-1 |(m1 – m2)/(1 + m1 m2)|

 

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

              If m1 (or m2) is infinity the angle is given by θ = |π/2 – θ1| where

              θ1 = tan-1 m2 (or tan-1 m1)

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

                                     Φ = Ψ1 - Ψ2

                                     ⇒ tan Φ = tan (Ψ1 - Ψ2)

                                                = (tan Ψ1 – tan Ψ2)/(1 + tan Ψ1 tan Ψ2 ),        where tan Ψ1 = f'(x1) and tan Ψ2 = g'(x1).

Angle between two curves

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

                tanΨ1 tanΨ2 = -1.

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

                tanΨ1 = tanΨ2.

 

Illustration:

Show that the curves ax2 + by2 = 1 and cx2 + dy2 = 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 (x1, y1). Then

ax12 + by12 = 1                                                           ...... (1)

and cx12 + dy12 = 1.                                                     ...... (2)

From (1) and (2), we get

(a - c)x12 + (b - d)y12 = 0.                                           ...... (3)

Slope of the tangent to the curve ax2 + by2 = 1, at (x1, y1) is given by

tan Ψ1 = [dy/dx](x1,y1) = -ax1/by1 .

Slope of the tangent to the curve cx2 + dy2 = 1 at (x1, y1) is given by

tan Ψ2 = [dy/dx](x1,y1) = -cx1/dy1. If the two curves cut orthogonally, we must have,

(-ax1/by1)(-cx1/dy1) = -1 ⇒ acx12 + bdy12 = 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 y2 = 4ax at points where x = a are at right angles to each other.

Solution 2:

y2 = 4ax

When x = a

y2 = 4a2

⇒ y = ± 2a

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

Slope of the tangent of the curve y2 = 4ax is

y dy/dx = 4a

dy/dx = 2a/y

for y = 2a, dy/dx = 1 = m1

for y = -2a dy/dx = - 1 = m2

m1m2 = -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.

Length of tangent and normal

If 'P1' be the projection of the point P on the x-axis then TP1 is called the sub-tangent (projection of line segment PT on the x-axis) and NP1 is called the sub normal (projection of line segment PN on the x-axis).

Tangent:

TP = PP1 cosec θ = y √(1 + cot2θ)

                            = \left | \frac{y\sqrt{1+(dy/dx)^{2}}}{dy/dx} \right |

Subtangent:

TP1 = PP1 cot θ

       = \left | \frac{y}{dy/dx} \right |

Normal:

PN = PP1 sec θ = y √(1 + tan2θ)

                         = \left | y\sqrt{1+\left ( dy/dx \right )^{2}} \right |

Subnormal:

NP1 = PP1 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 y2 - 16x – 8y = 0 are

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

Solution: Given curve is y2 - 16x – 8y = 0                       ….… (1)

Let P = (14, 7)

Then equation (1) can be written as y2 – 8y = 16x

or y2 – 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+4t2, 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 (4t2 – 1, 4 + 8t), dy/dx = 1/t

Hence, the equation of normal at (-1 + 4t2, 4+8t) is

y - 4 – 8t = – t(x + 1 – 4t2)

or tx + y – 4 – 8t + t – 4t3 = 0 ….….. (3)

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

14t + 7 – 4t3 – 7t – 4 = 0

or 4t3 – 7t -3 = 0

(t + 1)(4t2 – 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 x2 + y2 – 2x – 4y + 1 = 0 which are parallel to the x-axis.

Solution: The equation of the curve is x2 + y2 – 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, 12 + y2 – 2(1) – 4y + 1 = 0

So, y2 – 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) (dy1/dx)(dy2/dx) = -1

(b) (dy1/dx)(dy2/dx) = 1

(c) (dy1/dx)(dy2/dx) = 0

(d) (dy1/dx)2 = -1

Q2. If θ is the angle between y = x2 and 6y = 7-x3 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 + x2 = 7, a > 0 and x3 = 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 (x1,y1) 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


Q1. Q2. Q3. Q4. Q5.
(a) (d) (b) (c) (a)


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 JEEAIEEEDCEEAMCET and other engineering entrance examinations. 

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