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- Journal of Science and Arts Year 14, No. 3(28), pp. 199-210, 2014
ORIGINAL PAPER
SOME NEW IDENTITIES ON THE CONIC SECTIONS
DAM VAN NHI 1, TRAN TRUNG TINH1, PHAM MINH PHUONG 2
_________________________________________________
Manuscript received: 15.08.2014; Accepted paper: 22.09.2014;
Published online: 30.09.2014.
Abstract. In mathematics, a conic section (or just conic) is a curve obtained as the
intersection of a cone (more precisely, a right circular conical surface) with a plane. In this
paper, we construct some new identities and proposed the concept of the power of a point
with respect to a conic.
Keywords: Conic section, identity, power line, power of a point.
2010 Mathematics Subject Classification: 26D05, 26D15, 51M16.
1. THE ECCENTRICITY OF CONIC SECTION
Definition 1.1. A conic section (or conic) is a curve in which, a plane, not passing
through the cone's vertex, intersects a cone.
Conics possess a number of properties, one of them consisting in the following result.
Proposition 1.2. [2] Each conic section, except for a circle, is a plane locus of points
the ratio of whose distances from a fixed point F and a fixed line d is constant. The point F
is called the focus of conic, the line d its directrix.
Proof: Let ( ) be the curve in which the plane ( P ) intersects a cone. We inscribe a
sphere in the cone, which touches the plane ( P ) at the point F . Let (ω ) be the plane
containing the circle along which the sphere touches the cone. We take an arbitrary point
M ∈ ( ) and draw through it a generator of the cone, and denote by B the point of its
intersection with the plane (ω ). We then drop a perpendicular from M to the line d of
intrsection of the planes ( P ) and (ω ), example: MA ⊥ d . We obtain FM = BM because
they are the tangents to the sphere drawn from one point. Further, if we denote by h the
h h
distance of M from the plane (ω ), then AM = , BM = , where α is the angle
sin α sin β
between the planes (ω ) and ( P ) and β is the angle between the generator of the cone and
AM AM sin β
the (ω ). Hence it follows that = = .
FM BM sin α
AM sin β
=
Thus, the ratio λ = does not depend on the point M .
FM sin α
1
Hanoi National University of Education, Cau Giay, Hanoi, Vietnam. E-mail: tinhtckh@gmail.com.
2
Hanoi National University of Education, High School for Gifted Students, Hanoi, Vietnam.
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We note that if λ < 1 then ( ) is an ellipse; if λ = 1 then ( ) is a parabola and if λ > 1
then ( ) is a hypebol. The number λ is called the eccentricity of the conic section.
Let us now pass over to rectangular Cartesian coordinates Oxy in the plane ( P ),
where F (0,0) and d : x = p. Suppose that M ( x, y ). Then AM = λ FM if and only if
(1 − λ 2 ) x 2 − 2 pλ 2 x + y 2 − p 2λ 2 =
0.
p
(1) If λ = 1 and by putting − x by x then we obtain the canonical equation of the
2
p p
parabola ( P ) : y 2 = 2 px and F ( ,0), d : x = − .
2 2
pλ 2 2 p 2λ 2 pλ 2
(2) If λ < 1 then (1 − λ )( x + 2
) + y = 2 . By putting for brevity x + 2
2
1− λ2 1− λ 1− λ
pλ
2 2
pλ
2 2
by x and a 2 = , b2 = we get the canonical equation for the ellipse
(1 − λ )
2 2
1− λ2
x2 y2 pλ 2 pλ 2 p
1 and F (
(E) : 2 + 2 = ,0), d : x =
p+ = 2.
a b 1− λ 2
1− λ 2
1− λ
pλ 2 2 p 2λ 2 pλ 2
(3) If λ > 1 then (1 − λ 2 )( x + ) + y 2
= . By putting for brevity x +
1− λ2 1− λ2 1− λ2
p 2λ 2 p 2λ 2
by x and a = 2
, −b = 2 we get the canonical equation for the hypebol
2
(1 − λ 2 ) 2 1− λ
x2 y2 pλ 2 p
(H ) : − =
1 and F ( ,0), d : x = .
a 2
b 2
1− λ 2
1− λ2
2. PARAMETRIZATION AND POWER
x = pt 2
Proposition 2.1. The parabol ( P= ) : y 2 4 px, p ≠ 0, is parameterized by .
y = 2 pt.
Let the stright line d : x = k ( y − v ) + u be passed through by the point N (u, v ) which
intersected ( P ) at the points A and B . With the point M (1, k ) belong to the stright line
NA. NB NA. NB
x = 1 we obtain the identity 2
= v 2 − 4 pu. The ratio is called the {\rm power}
OM OM 2
of the point N with respect to parabol ( P ).
Proof: The coordinates of A, B are the solutions of consider
x = k ( y − v) + u
4 px = y .
2
Consider the system of equations:
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x − u= k ( y − v )
2
y − 4 pky + 4 pkv − 4 pu =
0.
Let y1 , y2 are the solutions of equation y 2 − 4 pky + 4 pkv − 4 pu = 0. Inaddition, we
have A( x1= k ( y1 − v ) + u, y1 ), B( x2= k ( y2 − v ) + u, y2 ).
NA. NB
Thus, we have 2
= | ( y1 − v )( y2 − v ) |= | v 2 − 4 pu | .
OM
NA. NB
Hence, we have identities = v 2 − 4 pu. Because of I ( p,0) , the power of a
OM 2
focus point I with respect to the Parabol ( P ) is −4 p 2 .
Exercise 2.2. Constructing power lines of two parabols.
Proposition 2.3. The circle (C ) : x 2 + y 2 =
1 is a rational planar graphs in ,
2t 1 − t2 2t
parameterized =
by x (t ) = , y (t ) , provided that =
x(∞) lim= 0,
1 + t2 1 + t2 t →∞ 1 + t 2
1− t2
y (∞) =lim =−1 . The equation of the tangent line At to (C ) at a point A( x0 , y0 ) ∈ (C )
t →∞ 1 + t 2
is xx0 + yy0 = 1.
Proof: Equation of a line ( d ) through point (0;1) ∈ (C ) with slope −t is
2t 1 − t 2
( d ) : y =−tx + 1. The line ( d ) meet (C ) at the points (0;1) and At ( , ).
1 + t2 1 + t2
2t 1 − t 2
The point ( , ) could be anywhere in (C ) except (0; −1). Provided that
1 + t2 1 + t2
2t 1 − t2
=
x ( ∞) lim= =
0, y ( ∞) lim= −1, At could be (0; −1).
t →∞ 1 + t 2 t →∞ 1 + t 2
Due to A( x0 , y0 ) ∈ (C ) , we have x02 + y02 =1.
Thus, At : 2 x0 ( x − x0 ) + 2 y0 ( y − y0 ) =
0 , in other words, At : xx0 + yy0 =
1.
Proposition 2.4. Given (C ) : x 2 + y 2 =
R 2 . Provided that N (u, v ) and a line with slope
k = tan α through point N , meet (C ) at A and B. Consider the point M ( R sin α , R cos α ) in
u2 v2
(C ) , we have NA. NB = u 2 + v 2 − R 2 , in other words, NA. NB.OM=
2
R4 ( + − 1). We
R2 R2
say that NA. NB.OM 2 is the power of a point N with respect to circle (C ), review
Proposition 2.6.
Proof: We could easily obtain this result.
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x2 y2
Proposition 2.5. The ellip ( E ) : + =
1 is parameterized by
a 2 b2
2a 2bt b3 − a 2bt 2
=x (t ) = , y ( t )
b 2 + a 2t 2 b2 + a 2t 2
with convention
2a 2bt
=
x ( ∞ ) lim= 0
t →∞ b 2 + a 2 t 2
y ( ∞) =lim b − a bt =−b.
3 2 2
t →∞ b 2 + a 2t 2
Proof: The line ( d ) through (0; b) ∈ ( E ) with slope −t :
2a 2bt b3 − a 2bt 2
( d ) : y =−tx + b. ( d ) meets ( E ) at (0; b) and At ( , ). The point
b 2 + a 2t 2 b 2 + a 2t 2
2a 2bt b3 − a 2bt 2
At ( , ) through all points of ( E ), exept (0; −b). With convention that
b 2 + a 2t 2 b 2 + a 2t 2
2a 2bt b3 − a 2bt 2
x ( ∞) =lim 2 = 0, y ( ∞ ) =lim =−b
t →∞ b + a 2 t 2 t →∞ b 2 + a 2 t 2
x02 y02
deduce At through (0; −b). \par\noindent Since A( x0 , y0 ) ∈ ( E ) we have 2 + 2 = 1.
a b
2x 2y xx yy
Hence At : 20 ( x − x0 ) + 20 ( y − y0 ) = 0 or At : 20 + 20 = 1.
a b a b
x2 y2
Proposition 2.6. The ellip ( E ) : + = 1 with a, b ∈ and the focus points F ( c,0),
a 2 b2
F ′( −c,0). Construct a line through the focus point F , meet ellip ( E ) at A and B . We have
1 1 2a
(1) + = .
FA FB b2
2b 2
(2) Minimum value of AB is .
a
(3) Assuming N (u, v ) and a line through N with slope k = tan α meet ( E ) at C and
u2 v2
D . With the point M ( a sin α , b cos α ) ∈ ( E ) we have NC. ND.OM
= 2
a 2b 2 ( + − 1).
a 2 b2
We call NC. ND.OM 2 is the power of the point N respect to ellip ( E ).
(4) The power of the focus point F respect to ellip ( E ) FA.FB.OM 2 is −b4 .
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1 1 π
Proof: (1) We calculate =
T + . Assuming α = ∠xFA ≤ , r = FA and
FA FB 2
A( x1 , y1 ). Draw AP ⊥ Ox. We have FP = r cos α and x-axis of A : x1= c + r cos α . Thus, we
have
FA + F ′A = 2a
2
FA − F ′A = ( x1 − c ) 2 + y12 − ( x1 + c ) 2 − y12 =
−4cx1.
2
We deduce
r + F ′A =
2a
cx cx
r − F ′A =
−2 1
a and
r= a − 1 .
a
b2
We have a =ra − cx1 =a − c( c + r cos α ) or FA= r=
2 2
.
a + c cos α
b2 1 1 2a
Similary, we have FB = . We deduce T = + = 2.
a − c cos α FA FB b
b2 b2 2ab2
(2) From AB = FA + FB = + = we deduce
a + c cos α a − c cos α a 2 − c 2 cos2 α
2ab2 2b2 2b2 π
AB ≥ 2 = . Thus, the minimum value of AB is , equality holds if α = or
a a a 2
FA ⊥ Ox.
(3) The Equation NC : y = k ( x − u ) + v or =
y kx + h with h= v − ku. Coodinates of
= y kx + h
2
C and D are solutions of x y2
a 2 + b 2 =
1.
Let x1 , x2 are two solutions of equation (b2 + a 2k 2 ) x 2 + 2h ka 2 x + a 2h 2 − a 2b2 =
0.
Thus, we have C ( x1 , y1= k ( x1 − u ) + v ) and D ( x2 , y2= k ( x2 − u ) + v ) .
Deduce NC. ND =
| (u − x1 )(u − x2 ) | (1 + k 2 ).
From (b2 + a 2k 2 ) x 2 + 2h ka 2 x + a 2h 2 − a 2b2 =(b2 + a 2k 2 )( x − x1 )( x − x2 ) we have
(b2 + a 2k 2 )(u − x1 )(u − x2 ) = b2u 2 + a 2v 2 − a 2b 2 .
1+ k2
Deduce NC. ND =| (u − x1 )(u − x2 ) | (1 + k ) =| b u + a v − a b | 2
2 2 2 2 2 2 2
.
b + a 2k 2
u2 v2
Conclude that NC. ND (b 2 cos2 α + a 2 sin= 2
α ) a 2b2 ( 2 + 2 − 1).
a b
c 2 02
(4) From c= a − b , deduce FA.FB.OM =
2 2 2
a b ( 2 + 2 − 1) =−b 4 .
2 2 2
a b
Exercise 2.7. Constructing power line of two ellipse.
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x2 y2
Proposition 2.8. Hypebol (H): − = 1 with a, b ∈ is a rational planar graphs in
a 2 b2
a + a bt2 2
2b 2 t
, parameterized
= by x (t ) = , y ( t ) , convention that
1 − b 2t 2 1 − b 2t 2
a + ab 2t 2 2b 2t
x(∞) =lim =
=− a , y ( ∞ ) lim= 0.
t →∞ 1 − b 2t 2 t →∞ 1 − b 2t 2
Proof: The line ( d ) through ( a;0) ∈ ( H ) with slope at: (d):x=a(ty+1). ( d ) meets
a + a bt2 2
2b 2 t a + a bt2 2
2b 2 t
( H ) at ( a;0) and At ( , ) . The point A ( , ) through all
1 − b 2t 2 1 − b 2t 2 1 − b 2t 2 1 − b 2t 2
t
points of ( H ), exept ( −a;0). With convention that
a + a bt2 2
2b 2 t
x ( ∞) =lim =− a , y ( ∞ ) =lim =0
t →∞ 1 − b 2t 2 t →∞ 1 − b 2t 2
deduce At through ( −a;0).
x2 y2
1 with a, b ∈ and the focus points
Proposition 2.9. Hypebol ( H ) : 2 − 2 =
a b
F ( c,0), F ′( −c,0). Consider:
(1) The line d with slope k through the focus point F1 meet Hypebol ( H ) at A and
ab abk F1 A.F1B b2
B . With M ( , ) ∈ (H ) we have = − , called power of
| b2 − a 2k 2 | | b2 − a 2k 2 | OM 2 a2
the point F1 with respect to ( H ).
(2) Suppose that N (u, v ) and a line with slope k through N meet ( H ) at C and D .
2
NC. ND 2 2 u v2 NC. ND
We have = 2
a b ( 2 − 2 − 1). The ratio is called the power of the point
OM a b OM 2
N relatvie to hypebol ( H ).
Proof: (1) From the focus point F1 of hypebol H , construct a line d =
: y k ( x − c)
meet ( H ) at A and B . We have the coordinates of A and B are solutions of
x2 y2 ( X + c)2 k 2 X 2
2− 2 = 1 − 2 = 1
a b or a2 b
=
y k ( x − c) y= kX , X= x − c.
The equation (b2 − a 2k 2 ) X 2 + 2b2 cX + b4 =
0 have two solutions X 1 , X 2 with
b4
X1X 2 = .
b2 − a 2k 2
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b4 (1 + k 2 ) b2 a 2b2 (1 + k 2 )
=
So F1 A.F1B = . .
| b2 − a 2k 2 | a 2 | b2 − a 2k 2 |
ab abk F1 A.F1B b2
With M ( , ) ∈ ( H ) we have = − .
| b2 − a 2k 2 | | b2 − a 2k 2 | OM 2 a2
(2) Similary, from proposition 2.6 we have the proof.
Exercise 2.10. Constructing the power line of two hypebols.
3. SOME IDENTITIES FOR THE CONIC SECTIONS
We proceed now to establish the fundamental identities for conic sections.
Proposition 3.1. Let A1 , A2 , A3 , A4 be the points belong to parabola ( P ) : y = ax 2 with
coordinates ( x1 , ax12 ), ( x2 , ax22 ), ( x3 , ax32 ) and ( x4 , ax42 ), respectivelly, where
x1 < x2 < x3 < x4 , with 6 following points
M 12 (a ( x2 + x1 ),1), M 23 (a ( x2 + x3 ),1), M 34 (a ( x3 + x4 ),1),
M 41 (a ( x4 + x1 ),1), M 13 (a ( x1 + x3 ),1), M 24 (a ( x2 + x4 ),1)
of the line d : y = 1, we have the following identities:
A1 A2 AA AA A4 A1
(1) + 2 3 + 3 4 = .
OM 12 OM 23 OM 34 OM 41
A1 A2 A3 A4 AA AA A1 A3 A2 A4
(2) . + 4 1 . 2 3 = . .
OM 12 OM 34 OM 41 OM 23 OM 13 OM 24
Proof: (1) Direct computation shows that the relation
A1 A22 =( x2 − x1 ) 2 [1 + a 2 ( x2 + x1 ) 2 ].
A1 A2
Then = x2 − x1. By an argument similar, we have 6 following relations:
OM 12
A1 A2 AA AA
= x2 − x1 , 2 3= x3 − x2 , 3 4= x4 − x3 ,
OM 12 OM 23 OM 34
A4 A1 AA AA
= x4 − x1 , 1 3= x3 − x1 , 2 4= x4 − x2 .
OM 41 OM 13 OM 24
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A1 A2 AA AA AA
Hence + 2 3 + 3 4 = x2 − x1 + x3 − x2 + x4 − x3 = 4 1 .
OM 12 OM 23 OM 34 OM 41
(2) We have
A1 A2 A3 A4 AA AA
. + 4 1 . 2 3 = ( x2 − x1 )( x4 − x3 ) + ( x4 − x1 )( x3 − x2 )
OM 12 OM 34 OM 41 OM 23
A1 A3 A2 A4
= ( x4 − x2 )( x3 − x1 ) = . .
OM 13 OM 24
Proposition 3.2. Let A1 , A2 , …, An , M be n + 1 points belong to parabola
( P ) : y = ax 2
with coordinates 2
Ai (( xi , ax ))
i and M ( x0 , ax02 ), respectivelly, where
x1 < x2 < x3 < x4 < < xn < x0 . With the points I i ( i +1) (a ( xi + xi +1 ),1), J i (a ( x0 + xi ),1) belong
to the line y = 1, where n + 1 ≡ 1 and=i 1, 2, …, n, we have the following identities:
A1 A2 A2 A3 A A AA
(1) + + + n −1 n =n 1 .
OI12 OI 23 OI ( n −1) n OI n1
A1 A2 A2 A3 An −1 An An A1
OI1 OI 2 OI n −1 OI n
(2) + ++ = .
MA1 MA2 MA2 MA3 MAn −1 MAn MAn MA1
. . . .
OJ 1 OJ 2 OJ 2 OJ 3 OJ n −1 OJ n OJ n OJ 1
Ai Ai +1
Proof: (1) Arguing as in above proof, we get = xi +1 − xi for i =1, …, n − 1 and
OI i ( i +1)
An A1
= xn − x1. By the computation, it is easy to verify that
OI n1
A1 A2 A2 A3 A A AA
+ + + n −1 n = xn − x1 = n 1 .
OI12 OI 23 OI ( n −1) n OI n1
MAi
(2) There is = x0 − xi for=i 1, 2, …, n. By an easy computation it follows that
OJ i
the ratio:
Ai Ai +1
OI i ( i +1) xi +1 − xi 1 1
= = − .
MAi MAi +1 ( x0 − xi )( x0 − xi +1 ) x0 − xi +1 x0 − xi
.
OJ i OJ i +1
A1 A2 A2 A3 An −1 An An A1
OI1 OI 2 OI n −1 1 1 OI n
Hence + ++ = − = .
MA1 MA2 MA2 MA3 MAn −1 MAn x0 − xn x0 − x1 MAn MA1
. . . .
OJ 1 OJ 2 OJ 2 OJ 3 OJ n −1 OJ n OJ n OJ 1
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Definition 3.3. Let a and b be an arbitrary pair of real numbers such that ab > 0. A
transformation under which any point M ( x, y ) shifts to L( ax, by ) is called the
transformation N ab .
Clearly, under the transformation inverse N ab−1 , any point ( x, y ) is sent into the point
( x , y ).
a b
Proposition 3.4. Let A, B, C , D be 4 points with the coordinates ( a cos t1 , b sin t1 ),
( a cos t2 , b sin t2 ), ( a cos t3 , b sin t3 ), ( a cos t4 , b sin t4 ), where 0 < t1 < t2 < t3 < t4 < 2π , belong to
x2 y2 t +t t +t t +t
the ellipse 2
+ 2 = 1. With 12 points I ij (a tan j i , b) and M ij (a sin j i , b cos j i ),
a b 2 2 2
where i, j = 1, 2,3, 4, i < j, and choosing properly u, v, t ∈ {1, −1} we have the following
identities
AB BC CD DA
(1) +u +v +t = 0.
OI12 OI 23 OI 34 OI 41
AB DC BC DA AC DB
(2) . +u . +v . =
0.
OI12 OI 34 OI 23 OI14 OI13 OI 24
AB CD DA CB AC BD
(3) +u +v =
0.
OM 12 OM 34 OM 41 OM 23 OM 13 OM 24
Proof: (1) Suppose that A( a cos t1 , b sin t1 ), B( a cos t2 , b sin t2 ).
t2 − t1 2 2 t2 + t1 t +t
=Then AB 2sin a sin + b2 cos2 2 1 or
2 2 2
AB
± = sin t2 − sin t1.
2 t2 + t1
2
a tan +b 2
2
AB
Thus, ± = sin t2 − sin t1. Upon simple computation, we get
OI12
AB BC CD DA
± ± = sin t4 − sin t1 = ± .
OI12 OI 23 OI 34 OI 41
AB BC CD DA
Then we obtain ± ± ± = 0.
OI12 OI 23 OI 34 OI 41
AB DC
(2) Since ± = sin t2 − sin t1 and ± = sin t4 − sin t3 we get
OI12 OI 34
AB DC
± . = (sin t2 − sin t1 )(sin t4 − sin t3 ).
OI12 OI 34
BC DA
Similar, there are the relations ± . = (sin t3 − sin t2 )(sin t4 − sin t1 ) and
OI 23 OI14
AC DB
± . = (sin t3 − sin t1 )(sin t4 − sin t2 ). Hence, there is the following relation
OI13 OI 24
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AB DC BC DA AC DB
. ± . ± . =
0.
OI12 OI 34 OI 23 OI14 OI13 OI 24
(3) follows from (2).
Lemma 3.5. Given the convex polygon A1 A2 … An M and it's circumcircle.
A1 A2 A2 A3 An −1 An AA
We have identity + ++ = n 1 . In case n = 3 ,
MA1.MA2 MA2 .MA3 MAn −1.MAn MAn .MA1
we get Ptolemy identity.
Proof: Assuming the circumcircle of a polygon A1 A2 … An M have radius R = 1.
Coodinates of A1 , A2 , …, An , M are z1 , z2 , …, zn , z respectivelly, where
= zk cos uk + i sin uk ,
=
k 1, 2, …, n, and= z cos u + i sin u (0 < u < u1 < u2 < … < un < 2π ) .
z1 − zn z1 − z2 z 2 − z3 zn −1 − zn
We have = + + +
( z − z1 )( z − zn ) ( z − z1 )( z − z2 ) ( z − z2 )( z − z3 ) ( z − zn −1 )( z − zn )
u2 − u1 − iu
z1 − z2 i 2sin e iA1 A2e − iu
and = = 2
( z − z1 )( z − z2 ) 4sin u1 − u sin u2 − u MA1.MA2
2 2
u3 − u2 − iu
z 2 − z3 i 2sin e iA2 A3e − iu
= = 2
( z − z2 )( z − z3 ) 4sin u2 − u sin u3 − u MA2 .MA3
2 2
=
u − un −1 − iu
zn −1 − zn i 2sin n e iAn −1 An e − iu
= = 2
( z − zn −1 )( z − zn ) 4sin un − u sin un −1 − u MAn −1.MAn
2 2
un − u1 − iu
z1 − zn i 2sin e iAn A1e − iu
= = 2 .
( z − z1 )( z − zn ) 4sin un − u sin u1 − u MAn .MA1
2 2
A1 A2 An −1 An AA
Hence ++ = n 1 .
MA1.MA2 MAn −1.MAn MAn .MA1
Proposition 3.6. Let A1 , A2 , …, An , M be n + 1 points belong to the ellipse
2 2
x y
(E) : 2
+ 2 = 1 with the coordinates ( a cos ti , b sin ti ) and M ( a cos t , b sin t ), respectivelly,
a b
t +t t +t
where 0 < t1 < t2 < t3 < t4 < < tn < t < 2π . With the points I i (a sin i +1 i , b cos i +1 i ),
2 2
t + ti t + ti
J i (a sin , b cos ), where n + 1 ≡ 1, i= 1, 2,…, n, and a proper choice ± we have the
2 2
following identity
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A1 A2 A2 A3 An −1 An An A1
OI1 OI 2 OI n −1 OI n
± ±± ± =
0.
MA1 MA2 MA2 MA3 MAn −1 MAn MAn MA1
. . . .
OJ 1 OJ 2 OJ 2 OJ 3 OJ n −1 OJ n OJ n OJ 1
Proof: Denote Bi = N ab−1 ( Ai ) for i= 1, …, n . By Lemma 3.5 and N = N ab−1 ( M ) we have
B1B2 B2 B3 Bn −1Bn Bn B1 AA
the identity ± ±± ± = 0. Because 1 2 = B1B2 , …,
NB1. NB2 NB2 . NB3 NBn −1. NBn NBn . NB1 OI1
An −1 An AA MA1 MA2 MAn
= Bn −1Bn , n 1 = Bn B1 and =NB1 , =NB2 , …, =NBn we get the identity
OI n −1 OI n OJ 1 OJ 2 OJ n
A1 A2 A2 A3 An −1 An An A1
OI1 OI 2 OI n −1 OI n
± ±± ± = 0.
MA1 MA2 MA2 MA3 MAn −1 MAn MAn MA1
. . . .
OJ 1 OJ 2 OJ 2 OJ 3 OJ n −1 OJ n OJ n OJ 1
Proposition 3.7. Let A, B, C , D be 4 points belong to hypebol
x2 y2
(H ) : 2 − 2 = 1
a b
with the coordinates
a + a bt2 2
2b 2 t a + a bu2 2
2b 2 u a + a bv2 2
2b 2 v a + a bz2 2
2b 2 z
A( , ), B ( , ) , C( , ), D ( , )
1 − b 2t 2 1 − b 2t 2 1 − b 2u 2 1 − b 2u 2 1 − b2v 2 1 − b2v 2 1 − b2 z 2 1 − b2 z 2
1 1 1 1
satisfies > > > . With the pints
1− b t 2 2
1− b u
2 2
1− b v2 2
1 − b2 z 2
1 + b2tu 1 + b2tv 1 + b2tz
I ab ( a, ), I ac ( a, ), I ad ( a, )
t+u t+v t+z
1 + b2uv 1 + b2uz 1 + b2vz
I bc ( a, ), I bd ( a, ), I cd ( a, )
u+v u+z v+z
we have
AB BC CD AD
(1) + + =.
OI ab OI bc OI cd OI ad
AB CD AD BC AC BD
(2) . + . = . .
OI ab OI cd OI ad OI bc OI ac OI bd
Proof: We have
( a + a bt a + a bu
2 2 2 2
2b 2 t 2b 2 u 2
AB 2 = − ) 2
+ ( − ).
1 − b 2t 2 1 − b 2u 2 1 − b 2t 2 1 − b 2u 2
1 + b2tu 2
2b 2 | t 2 − u 2 | a 2 + ( )
Thus, AB = t+u . By computation the relation
|1 − b2t 2 ||1 − b2u 2 |
AB 2b 2 | t 2 − u 2 | 1 1
= = 2( − )
OI ab |1 − b t ||1 − b u |
2 2 2 2
1 − b t 1 − b 2u 2
2 2
ISSN: 1844 – 9581 Mathematics Section
- 210 Some new identities on the conic sections Dam Van Nhi et all
1 AC1 1 1
− = |= 2(
2| − )
1− b t 1− b v
2 2
OI ac2 2
1 − b t 1 − b2v 2
2 2
1 AD1 1 1
2| − = |= 2( − )
1− b t 1− b z
2 2
OI ad 2 2
1 − b t 1 − b2 z 2
2 2
1 BC1 1 1
2| − = =| 2( − )
1− b u 1− b v
2 2
OI bc2 2
1 − b u 1 − b2v 2
2 2
1 BD 1 1 1
2| − = =| 2( − )
1− b u 1− b z
2 2
OI bd2 2
1 − b u 1 − b2 z 2
2 2
1 CD 1 1 1
2| − = =| 2( − ).
1− b v 1− b z
2 2
OI cd2 2
1 − b v 1 − b2 z 2
2 2
AB CD AD BC AC BD
Hence, we obtain . + . = . and (2). We have
OI ab OI cd OI ad OI bc OI ac OI bd
AB 1 1
= 2( − ). The anothers relations are proved in an analogous fashion.
OI ab 1 − b t 1 − b 2u 2
2 2
AB BC CD AD
Then, we have the identity + + = and (1).
OI ab OI bc OI cd OI ad
Proposition 3.8. Let A1 , A2 , …, An , An +1 be n + 1 points belong to hypebol
x 2
y 2
a + a bt2 2
2b2ti
(H ) : − 2 =
1 with the coordinates Ai ( , i
), i = 1, 2, …, n + 1, and
a 2
b 1 − b t 1 − b2ti2
2 2
i
a + ab2t 2
M( ) satisfy the conditions 1 2 2 > 12 2 , where=i 1, 2,…, n, and the points
1− b t
2 2
1 − b ti 1 − b ti +1
1 + b 2tr t s
I r ,s (a , ), r, s = 1, 2,…, n + 1. Then, we have the following identities:
tr + t s
n
Ai Ai +1 A1 An +1
(1) ∑ OI
i =1
=
OI1,n +1
.
i ,i +1
Ai Ai +1 An A1
n
OI i ,i +1 OI n ,1
(2) ∑ = .
i =1
Ai An +1 Ai +1 An +1 An An +1 A1 An +1
OI i ,n +1 OI i +1,n +1 OI n ,n +1 OI1,n +1
REFERENCES
[1] Mau, N.V., Nhi, D.V., Identities and coordinate method in Geometry, The National
University Publishing House, Hanoi, 2012.
[2] Pogorelov, A., Geometry, Mir Publishers, Moscow, 1987.
www.josa.ro Mathematics Section
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