Tangent and normal

APPLICATIONS OF DERIVATIVES
TANGENT AND NORMAL
mathews.suman@gmail.com

Given a curve y = f(x), slope of the tangent to the curve at ),( 11 yx
is
),( 11
)( yx
dx
dy
Slope of the normal to the curve at ),( 11 yx is
)( 1,1
)(
1
yx
dx
dy
Equation of the tangent : )()( 1)(1 1,1
xx
dx
dy
yy yx 
Equation of the normal:
)(
)(
1
1
),(
1
1
xx
dx
dy
yy
yx


Find the point on the curve
123 2
 xxy at which the slope
of the tangent is 4
26  x
dx
dy
6x -2 = 4
6x = 6
x = 1
y = 2
Required point =( 1,2)

Find the points on the curve 1322
 yx the tangent at each
one of which is parallel to the line 2x + 3 y = 7
3
2
3
7
3
2
723
732






m
xy
xy
yx
y
x
dx
dy
x
dx
dy
y
dx
dy
yx
dx
dy
yx
yx





0
022
1322
3
2


y
x
yx
3
2


𝑥2
+ 𝑦2
= 13
4
9
𝑦2 + 𝑦2 = 13
13𝑦2
9
= 13
y=±3
x=
2
3
3 = 2 One point of intersection = (2,3)
𝑥 =
2
3
−3 = −2 Other point of intersection = (-2,3)

The acute angle between the 2 curves is the angle between the tangents
to the 2 curves at their point of intersection.
21
21
1
tan
mm
mm



121 mm The curves cut each other orthogonally
21 mm  The curves touch each other

Find the condition that the curves
2
2 yx 
and 2xy =k intersect orthogonally
Solving both equations simultaneously
Point of intersection = ),
2
( 3/1
3/2
k
k
ky 3
3/1
ky 
2
2
y
x 

For the 1st curve 22 
dx
dy
y
13/1
11
m
kydx
dy

For the 2nd curve, kxy 2
x
k
y
2

2
2x
k
dx
dy 

3/4
2
4
k
k

23/1
2
m
k




121 mm
1
2
3/13/1


kk
23/2
k
82
k

Prove that the curves xy = 4 and 822
 yx touch each other
We first find the point of intersection of the 2 curves
y
x
4
 8
16 2
2
 y
y
24
816 yy  0168 24
 yy
0)4( 22
y 42
y 2y
Point of intersection = (2,2) (-2,-2)
When y = 2, x=2 when y = -2, x= -2

x
y
4

1m at (2,2) is -1
22
8 xy 
x
dx
dy
y 22 
y
x
dx
dy

  1)2,2(2 m
Since 21 mm  the curves touch each other
2
4
xdx
dy

It can also be shown that at (-2,-2), the curves touch each other.

Tangent and normal

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