Cryptography Reference
In-Depth Information
If
x
1
=
x
2
,then
L
is vertical. We'll treat this case later, so let's assume that
x
1
=
x
2
. The equation of
L
is then
y
=
m
(
x
−
x
1
)+
y
1
.
To find the intersection with
E
, substitute to get
x
1
)+
y
1
)
2
=
x
3
+
Ax
+
B.
(
m
(
x
−
This can be rearranged to the form
0=
x
3
m
2
x
2
+
−
···
.
The three roots of this cubic correspond to the three points of intersection of
L
with
E
. Generally, solving a cubic is not easy, but in the present case we
already know two of the roots, namely
x
1
and
x
2
,since
P
1
and
P
2
are points
on both
L
and
E
. Therefore, we could factor the cubic to obtain the third
value of
x
. But there is an easier way. As in Chapter 1, if we have a cubic
polynomial
x
3
+
ax
2
+
bx
+
c
with roots
r, s, t
,then
x
3
+
ax
2
+
bx
+
c
=(
x
t
)=
x
3
(
r
+
s
+
t
)
x
2
+
−
r
)(
x
−
s
)(
x
−
−
···
.
Therefore,
r
+
s
+
t
=
−a.
If we know two roots
r, s
, then we can recover the third as
t
=
−
a
−
r
−
s
.
In our case, we obtain
x
=
m
2
− x
1
− x
2
and
y
=
m
(
x − x
1
)+
y
1
.
Now, reflect across the
x
-axis to obtain the point
P
3
=(
x
3
,y
3
):
x
3
=
m
2
− x
1
− x
2
,
y
3
=
m
(
x
1
− x
3
)
− y
1
.
=
y
2
, the line through
P
1
and
P
2
is a vertical
line, which therefore intersects
E
in
Inthecasethat
x
1
=
x
2
but
y
1
∞
. Reflecting
∞
across the
x
-axis yields
the same point
at both the top and the bottom of
the
y
-axis). Therefore, in this case
P
1
+
P
2
=
∞
(this is why we put
∞
.
Now consider the case where
P
1
=
P
2
=(
x
1
,y
1
). When two points on
a curve are very close to each other, the line through them approximates a
tangent line. Therefore, when the two points coincide, we take the line
L
through them to be the tangent line. Implicit differentiation allows us to find
the slope
m
of
L
:
∞
=
3
x
1
+
A
2
y
1
2
y
dy
dx
dy
dx
=3
x
2
+
A,
so
m
=
.
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