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Parabola, Ellipse, and Hyperbola fee Lees poole ANUS Hyperbola is a conic © ECcentricity ; erbola is a Ticity is wie iy 8reater than (b) Conjugate Axis: The line segment B’B between yal the two points 8” = (0, -b) and B = (0, 6) is called . as the Conjugate Axis of the Hyperbola. aandard Equation and The Transverse axis and the Conjugate axis of the tion s) hyperbola are together called the Principal axes of the — hyperbola. jd equation of the hyperbola is 6. Focal Property: The difference of the focal distances , 2 of any point on the hyperbola is constant and equal to 5-3 ah transverse axis, ie. ||PS|-|PS‘l| = 2a. The distance a b SS’ = focal length. Pea(e-}) 7. Focal distance: Distance of any point P (x, y) on ere Hyperbola from foci PS = ex—a and PS’ =ex+a. @e=at bh ‘ 2 i a Find the equation of the hyperbola whose directrix is ¥ 2x +y=1, focus (1, 2) and eccentricity v3. SOLUTION Perpendicular from P on the directrix. y | Let P (x, y) be any point on the hyperbola and PM is | Then by definition x = (ae, -b’/a) SP =e PM 4 ; => (SP) =e (PM} j 2 : + \2 Qx+ : =1 + ( Souivante avs ) = cp g-2=af ary } { Transverse axis } 4 Pe PP De — } \. Foci: S= (ae, 0) and S’ = (-ae, 0). = Str 2e-4y+5) F | 2 Equations of directrices: x =“ andx =—4- eee ee an “iq iy ca z = Te 2y+ xy 2+ Ly — 22 ! : 4 Vertices: 4 = (a, 0) and A’ = (-a, 0). =0 I 7 4, Latus rectum: which is the required hyperbola. (a) Equation: x =+ ae 2 257 Conjugate axis ) Piikeneth = a = axis = 2a (e?— 1) =2e y ofthe by | The eccentricity of the hyperbola 4x?- 9y?— 8x = 32 is (distance from focus to directrix) SOLUTION } j 2 —p? -b7 ~9y?_ 8x = (© Ends: { ae, 2 \,| ae, =) ; (-. = } 4x°— 9y?— 8x = 32 a a = 4 (x- 1¥-97 =36 Sa Transverse Axis: The line segment A’A of length @-y ¥ | 2a in which the foci S” and S both lie is called the rs) wos a] 9 4 Transverse Axis of the Hyperbola. 
