8.3 THE HYPERBOLA OBJECTIVES

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1 8.3 THE HYPERBOLA OBJECTIVES 1. Define Hperol. Find the Stndrd Form of the Eqution of Hperol 3. Find the Trnsverse Ais 4. Find the Eentriit of Hperol 5. Find the Asmptotes of Hperol 6. Grph Hperol

2 HPERBOLAS WITH CENTER AT ORIGIN ( 0,0) It ws shown in the introdution of hpter 8, tht hperol is the urve tht ours t the intersetion of one with two nppes nd plne. The definition of hperol is similr to tht of n ellipse. The onl hnge is tht insted of using the sum of distnes from two fied points, we use the differene. Definition of n Hperol A hperol is the set of ll points in plne, the differene of whose distnes from two fied points( the foi) in the plne is positive onstnt P(, ) F,0 V 0,0 V F,0 ( ) ( ) ( ) Figure To otin simple eqution for hperol, let us hoose the foi to e t F(,0) nd F (,0) where > 0, nd denote the onstnt distne where > 0. The midpoint of the line segment FF (the origin(0,0)) is lled the enter of the hperol. Aording to the definition of hperol, if the point P( is, ) on the hperol (see Figure 5.3.1), then we must hve = ( ) ( )

3 From whih we get (see the tetook) where > > 0 Sine is positive, we m reple it nother positive numer,. Thus, where = is the stndrd form of the eqution of hperol entered t the origin with foi on the is. Hperol whose Trnsverse Ais in on the -is: F,0 V,0 V,0 F W ( 0, ) ( ) ( ) ( ) (,0) W ( 0, ) Properties: Appling tests for smmetr, we see tht the hperol is smmetri with respet to oth es nd the origin( Wh?). Letting = 0 in the stndrd form gives = =± Figure 5.3.

4 V(,0) V (,0) thus the -interepts of hperol re nd s shown in Figure V nd V re lled the verties of the hperol. The line segment VV is lled the trnsverse is of the hperol. The grph hs no -interepts, sine the eq ution = 1hs the nonrel solutions = ± i. However, the points W( 0, ) nd W ( 0, ) re ver importnt, s we will see. The line segment WW is lled the onjugte is of the hperol. Now, s ou see in Figure nd Figure 5.3., tht the grph of hperol hs two rnhes. The two rnhes looks like prols, ut the re not prols. The rnhes of the hperol shown in Figure 5.3. get loser nd loser to the dshed lines, lled the smptotes, ut the never interset them. The smptotes re used s guidelines in skething hperol. The smptotes re found etending the digonls of the fundmentl retngle, shown in Figure It is ler from Figure tht the smptote with positive slope psses through origin, while the other smptote psses through origin nd the point nd the point ( ) (, ). Therefore the lines = nd = re smptotes for the hperol

5 EXAMPLE 1 Sketh nd desrie the hperol 9 16 Solution ( ) W ( 0,4) F 5, 0 V 3,0 V 3,0 ( ) ( ) F ( 5,0) Compre with, we see 9 16 tht = 9 nd = 16 Hene = + = 5. Figure W ( 0, 4) ( 3,0 ) nd V ( 3,0) Thus the eqution represents hperol with the following properties: Verties t V Center t (0,0) Trnsverse is on the -is. Conjugte is on the -is with endpoints W ( 0,4) nd W ( 0, 4). Foi t F ( 5,0) nd F ( 5,0) 4 4 The equtions of the smptotes re = nd =. 3 3

6 ( ± 3, 4) ( 3, ± 4) Now drw the fundmentl retngle whose verties re t nd whose digonl lie on the smptotes of the hperol s shown in Figure A rough sketh of the hperol n e drwn s in Figure Hperol with Trnsverse Ais on the -is endpoints W(,0) nd W (,0) Foi t F( 0, ) nd F ( 0, ) The eqution of the smptotes re Similrl, it n e shown tht if we tke the foi of the hperol on the -is, we otin the eqution with the following properties: Verties t V( 0, ) nd V ( 0, ) Trnsverse is on the V 0, nd -is with endpoints ( ) V ( 0, ) Conjugte is on the -is with where = + = nd = s shown in Figure

7 EXAMPLE Sketh nd desrie the hperol Solution If we follow the sme steps s in Emple 1, we get the properties of this hperol s follows, nd its grph s shown in Figure Properties of the given hperol The trnsverse is lies on the is (Wh?) nd with enter t origin. = 9 = V 0,3 nd 3 verties t ( ) V ( 0, 3) = 16 = 4 the endpoints of the onjugte is re t W 4,0 nd W ( 4,0) ( ) = + = 5 = 5 foi t 0,5 F 0, 5 F ( ) nd ( ) The eqution of the smptotes re nd = 3 4 = 3 4 F ( 0,5) V ( 0,3) W ( 4,0) W ( 4,0) V ( 0, 3) F ( 0, 5) Figure = = 4

8 THE ECCENTRICITY OF A HYPERBOLA As in the se of the ellipse, the eentriit e of hperol is given Sine >, then e > 1. e = Atull, the grph of hperol n e ver wide or ver nrrow: nrrow hperols hve e ner 1 wide hperols hs lrge e s shown in Figure e lose to 1 lrge e Figure 5.3.6

9 EXAMPLE 3 The hperols, given in emples 1 nd hve the sme eentriit e = =. 3 HYPERBOLA WITH CENTER AT ( hk, ) Like irles nd ellipses, hperols m e entered t n point in the plne. To get the eqution of hperol entered t ( hk, ), reple h nd k in the eqution of the hperol entered t origin s shown in the following tles nd Figures. In the two ses tht re given elow: Move units to the right to find V nd units to the left to find V (endpoints of the trnsverse is) Move units to the right to find F nd units to the left to find F Move units to up nd units to down to find the endpoints of the onjugte is.

10 Trnsverse Ais Prllel to the ( hk, ) F F V V is Eqution: ( h) ( k) Verties: V h, k ( + ) nd V ( h, k) Figure Figure Foi: F h, k ( + ) nd F ( h, k) where, = + Asmptotes: k =± h ( ) Eentriit: e = Oserve tht the verties nd foi lie on the line = k while the onjugte is lies on the line = h

11 Trnsverse Ais Prllel to the F V V F Figure ( hk, ) is Eqution: ( k) ( h) Verties: V h, k ( + ) nd V ( h, k ) Foi: F h, k ( + ) nd F ( h, k ) where, = + Asmptotes: k =± h ( ) Eentriit: e = In this se the verties nd foi lie on the line = h while the onjugte is lies on the line = k

12 EXAMPLE 4 Show tht the eqution = 0 represents hperol. Then desrie the hperol nd sketh its grph. Solution The given eqution n e rewritten s: 4( + 4 ) ( 8 ) = 16 We now omplete the squres: 4( ) ( 8+ 16) = ( + ) ( 4) = 16 ( + ) ( 4) 4 16 Whih is the stndrd form of hperol with the following properties: enter t (,4) Trnsverse is is prllel to the is nd lies on the line = 4 = 4 = V ( +,4) = ( 0,4) nd V (,4) = ( 4,4) 16 4 (, 4 4) (,8) W (,4 4) = (,0), nd lies on the line = verties t = = endpoints of the onjugte is W + = nd

13 F ( + 5,4) F ( 5,4 ) = + = 0 = 5 foi t nd, nd lies on the line = 4 Asmptotes: k h nd Eentriit e= e= 5 The grph of this hperol is shown in Figure =± ( ) 4= ( + ) 4= ( + ) (,4) V V F F Figure 5.3.8

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