Tangent And Normal
Presented By:- Ramesh Singh Makar Presented To:- Mr. Sunil Bhardwaj
1. Definition
• Tangent
• Normal
2. Equation
• Tangent
• Normal
3. Length Of Tangent, Sub-Tangent, Length Of Normal, Sub-Normal
CONTENTS
Tangent :-A tangent is a line that touches a curve. A tangent meets or touches a
circle only at one point, whereas the tangent line can meet a curve at
more than one point.
(1) (2)
 On the other hand, a line may meet the curve once, but still not be a tangent
The normal line to a curve at a particular point is the line
through that point and perpendicular to the tangent.
Normal :-
X
Y
p(x₁ ,y₁)
.
T
Normal
Tangent
Y = f(X)
o
Equation Of Tangent
 Let Y = f(X) be the given Curve and p(x₁ , y₁) the given point on curve . Let PT be the
Tangent at the point p(x₁ , y₁).
Then the slope of the tangent to the curve is equal to dy/dx at point p.
Slope of Tangent PT = dy/dx (x₁ , y₁)
= dy₁/dx₁
Equation of Tangent to the Curve at p(x₁ ,y₁)
Y-y₁ = dy₁/dx₁ (X-x₁)
This is required equation at point p.
.
P(x₁ , y₁)
Y = f(X)
Tangent
Y
X
·
T
O
𝑎2
+ 𝑏2
= 𝑐2
Equation Of Normal
 We Know that,
Normal is perpendicular to the Tangent at point p(x₁ , y₁) .
So…….
(slope of Tangent at p)(slope of Normal at p) = -1
m₁m₂ = -1 m₂ = -dx₁/dy₁
Required Equation of Tangent at point p(x₁ , y₁) Is……
Y-y₁ = (- dx₁/dy₁)(X-x₁)
Y = f(x)
Tangent
Normal
p(x₁ ,y₁)
o X
Y
Length of Tangent , Normal , Sub-
Tangent and Sub-Normal
X
Y
P(x,y)
Y =f(X)
G
T
o M
Let the Tangent and Normal at point p(x,y) to the Curve Y = f(X) meet the x-axis
at T and G respectively. And let PM with the ordinate through point P.
Here……
 PT is the Length of Tangent.
 PG is the Length of Normal.
 TM is the Length of Sub-Tangent .
MG is the Length of Sub-Normal.
1) Length of Tangent
In Triangle PMT
SinΨ = PM/PT = y/PT
PT = y/SinΨ = y CosecΨ
= y √(1+Cot²ψ)
Continue…….
PT = y√(1+1/tan²ψ)
= y√(1+tan²ψ)/tan²ψ
= y√{1+(dy/dx)²}/(dy/dx)
PT = y√1+(dx/dy)²
2) Length of Normal
In Triangle PMG
CosΨ = PM/PG = y/ PG
Continue…….
PG = y/Cosψ
PG = ySecψ
= y√(1+tan²ψ)
PG = y√1+(dx/dy)²
3) Length of Sub-Tangent
In Triangle PMT
tanψ = PM / TM
TM = y / (dy/dx)
4) Length of Sub-Normal
In Triangle PMG
tan ψ = MG/ PM
Y₁ = MG/PG
MG = yy₁
Ak.pdf

Ak.pdf

  • 1.
    Tangent And Normal PresentedBy:- Ramesh Singh Makar Presented To:- Mr. Sunil Bhardwaj
  • 2.
    1. Definition • Tangent •Normal 2. Equation • Tangent • Normal 3. Length Of Tangent, Sub-Tangent, Length Of Normal, Sub-Normal CONTENTS
  • 3.
    Tangent :-A tangentis a line that touches a curve. A tangent meets or touches a circle only at one point, whereas the tangent line can meet a curve at more than one point. (1) (2)  On the other hand, a line may meet the curve once, but still not be a tangent
  • 4.
    The normal lineto a curve at a particular point is the line through that point and perpendicular to the tangent. Normal :- X Y p(x₁ ,y₁) . T Normal Tangent Y = f(X) o
  • 5.
    Equation Of Tangent Let Y = f(X) be the given Curve and p(x₁ , y₁) the given point on curve . Let PT be the Tangent at the point p(x₁ , y₁). Then the slope of the tangent to the curve is equal to dy/dx at point p. Slope of Tangent PT = dy/dx (x₁ , y₁) = dy₁/dx₁ Equation of Tangent to the Curve at p(x₁ ,y₁) Y-y₁ = dy₁/dx₁ (X-x₁) This is required equation at point p. . P(x₁ , y₁) Y = f(X) Tangent Y X · T O 𝑎2 + 𝑏2 = 𝑐2
  • 6.
    Equation Of Normal We Know that, Normal is perpendicular to the Tangent at point p(x₁ , y₁) . So……. (slope of Tangent at p)(slope of Normal at p) = -1 m₁m₂ = -1 m₂ = -dx₁/dy₁ Required Equation of Tangent at point p(x₁ , y₁) Is…… Y-y₁ = (- dx₁/dy₁)(X-x₁) Y = f(x) Tangent Normal p(x₁ ,y₁) o X Y
  • 7.
    Length of Tangent, Normal , Sub- Tangent and Sub-Normal X Y P(x,y) Y =f(X) G T o M Let the Tangent and Normal at point p(x,y) to the Curve Y = f(X) meet the x-axis at T and G respectively. And let PM with the ordinate through point P.
  • 8.
    Here……  PT isthe Length of Tangent.  PG is the Length of Normal.  TM is the Length of Sub-Tangent . MG is the Length of Sub-Normal. 1) Length of Tangent In Triangle PMT SinΨ = PM/PT = y/PT PT = y/SinΨ = y CosecΨ = y √(1+Cot²ψ)
  • 9.
    Continue……. PT = y√(1+1/tan²ψ) =y√(1+tan²ψ)/tan²ψ = y√{1+(dy/dx)²}/(dy/dx) PT = y√1+(dx/dy)² 2) Length of Normal In Triangle PMG CosΨ = PM/PG = y/ PG
  • 10.
    Continue……. PG = y/Cosψ PG= ySecψ = y√(1+tan²ψ) PG = y√1+(dx/dy)² 3) Length of Sub-Tangent In Triangle PMT tanψ = PM / TM TM = y / (dy/dx)
  • 11.
    4) Length ofSub-Normal In Triangle PMG tan ψ = MG/ PM Y₁ = MG/PG MG = yy₁