Question

Find the equation of tangent and normal in the Cartesian form
​

Answer 1

Let the independent variable at
x
0
has the increment
Δ
x
.
The corresponding increment of the function
Δ
y
is expressed as

Δ
y
=
f
(
x
0
+
Δ
x
)
−
f
(
x
0
)
.
In Figure
1
,
the point
M
1
has the coordinates
(
x
0
+
Δ
x
,
y
0
+
Δ
y
)
.
We draw the secant
M
M
1
.
Its equation has the form

y
−
y
0
=
k
(
x
−
x
0
)
,
where
k
is the slope coefficient depending on the increment
Δ
x
and equal

k
=
k
(
Δ
x
)
=
Δ
y
Δ
x
.
When
Δ
x
decreases, the point
M
1
moves to the point
M
:

M
1
→
M
.
In the limit
Δ
x
→
0
the distance between the points
M
and
M
1
approaches zero. This follows from the continuity of the function
f
(
x
)
at
x
0
:
lim
Δ
x
→
0

Δ
y
=
0
,
⇒
lim
Δ
x
→
0

|
M
M
1
|
=
lim
Δ
x
→
0

√
(
Δ
x
)
2
+
(
Δ
y
)
2
=
0.
The limiting position of the secant
M
M
1
is just the tangent line to the graph of the function
y
=
f
(
x
)
at point
M
.
There are two kinds of tangent lines – oblique (slant) tangents and vertical tangents.

Definition
1
.
If there is a finite limit
lim
Δ
x
→
0

k
(
Δ
x
)
=
k
0
,
then the straight line given by the equation

y
−
y
0
=
k
(
x
−
x
0
)
,
is called the oblique (slant) tangent to the graph of the function
y
=
f
(
x
)
at the point
(
x
0
,
y
0
)
.
Definition
2
.
If the limit value of
k
as
Δ
x
→
0
is infinite:
lim
Δ
x
→
0

k
(
Δ
x
)
=
±
∞
,
then the straight line given by the equation

x
=
x
0
,