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In-Depth Information
∂
∂
f
X
∂
∂
f
Y
(
)
=
(
)
= 0
ab
,
ab
,
if and only if
∂
∂
F
X
∂
∂
F
Y
(
)
=
(
)
=
ab
,,
1
ab
,,
1
0
.
This means that there is no problem matching the zeros of the partials of f and F with
respect to X and Y. The only sticky part is ∂F/∂Z. However, Euler's formula (see
Theorem B.7.8) implies that
∂
∂
F
X
∂
∂
F
Y
∂
∂
F
Z
X
+
Y
+
Z
=
dF
,
where d is the degree of F. Therefore,
∂
∂
F
X
∂
∂
F
Y
(
)
=
(
)
=
(
)
=
Fab
,,
1
ab
,,
1
ab
,,
1
0
if and only if
∂
∂
F
X
∂
∂
F
Y
∂
∂
F
Z
(
)
=
(
)
=
(
)
=
ab
,,
1
ab
,,
1
ab
,,
1
0
.
Induction and a similar argument for higher partials finish the proof of the theorem.
10.6.12. Theorem.
If
p
is a regular point on a plane curve
C
defined by F(X,Y,Z),
then the equation of the tangent to
C
at
p
is
∂
∂
F
X
∂
∂
F
Y
∂
∂
F
Z
()
+
()
+
()
= 0.
X
p
Y
p
Z
p
Proof.
Let us use the same coordinate system and notation as in Theorem 10.6.11.
Then equation (10.40) shows that a tangent line is defined by
(
)
(
)
+
(
)
(
)
= 0
fabXa fabYb
,
-
,
-
,
X
Y
and so
X
Z
aFab
Y
Z
Ê
Ë
ˆ
¯
+
Ê
Ë
ˆ
¯
(
)
(
)
Fab
,,
1
-
,,
1
-
b
=
0
.
(10.45)
X
Y
But since F(a,b,1) = 0, the Euler formula implies that
(
)
=-
(
)
-
(
)
Fab
,,
1
aF ab
,,
1
bFab
,,.
1
(10.46)
Z
X
Y
