Some new identities on the conic sections

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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 "Some new identities on the conic sections", we construct some new identities and proposed the concept of the power of a point with respect to a conic.

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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. ISSN: 1844 – 9581 Mathematics Section
200 Some new identities on the conic sections Dam Van Nhi et all 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: www.josa.ro Mathematics Section
Some new identities on the conic sections Dam Van Nhi et all 201  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.  ISSN: 1844 – 9581 Mathematics Section
202 Some new identities on the conic sections Dam Van Nhi et all 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 . www.josa.ro Mathematics Section
Some new identities on the conic sections Dam Van Nhi et all 203 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. ISSN: 1844 – 9581 Mathematics Section
204 Some new identities on the conic sections Dam Van Nhi et all 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 www.josa.ro Mathematics Section
Some new identities on the conic sections Dam Van Nhi et all 205 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 ISSN: 1844 – 9581 Mathematics Section
206 Some new identities on the conic sections Dam Van Nhi et all 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 www.josa.ro Mathematics Section
Some new identities on the conic sections Dam Van Nhi et all 207 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 ISSN: 1844 – 9581 Mathematics Section
208 Some new identities on the conic sections Dam Van Nhi et all 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 www.josa.ro Mathematics Section
Some new identities on the conic sections Dam Van Nhi et all 209 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