Equation of Hyperbola

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Equation of Hyperbola

  1. 1. T- 1-855-694-8886 Email- info@iTutor.com T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
  2. 2. Hyperbola A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant. A hyperbola is a curve where the distances of any point from: – a fixed point (the focus), and – a fixed straight line (the directrix) are always in the same ratio. focus Directrix These distance are always in same ratio. focus Directrix © iTutor. 2000-2013. All Rights Reserved
  3. 3.  The two fixed points are called the foci of the hyperbola.  The mid-point of the line segment joining the foci is called the centre of the hyperbola.  The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis.  The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola. focus Directrix focus Directrix vertex vertex conjugate axis Transverse axis Centre The "asymptotes“ are not part of the hyperbola, but show where the curve would go if continued indefinitely in each of the four directions. © iTutor. 2000-2013. All Rights Reserved
  4. 4.  We denote the distance between the two foci by 2c, the distance between two vertices by 2a and we define the quantity b as Also 2b is the length of the conjugate axis To find the constant PF – PG: By taking the point P at A and B in the, we have BF – BG = AG – AF (by the definition of the hyperbola) BA + AF – BG = AB + BG – AF i.e., AF = BG , so that , BF – BG = BA + AF – BG = BA = 2a 22 acb 2c 2a a XX’ Y Y’ F GA B b c
  5. 5. Standard equation of Hyperbola • The equation of a hyperbola is simplest if the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis. – We will derive the equation for the hyperbola shown in with foci on the x-axis. Let F and G be the foci and O be the mid-point of the line segment FG. Let O be the origin and the line through O through G be the positive x-axis and that through F , as the negative x axis. The line through O to the x-axis be the y-axis. G(c , 0)F(-c , 0 ) P (x , y) x=-a x=a O XX’ Y Y’ © iTutor. 2000-2013. All Rights Reserved
  6. 6. Let the coordinates of F be (–c, 0) and G be (–c, 0) . Let P(x , y) be any point on the hyperbola such that the difference of the distances from P to the farther point minus the closer point be 2a. So given, PF – PG = 2a Using the distance formula, we have Squaring both side we get square binomials aycxycx 2)()( 2222 2222 )(2)( ycxaycxor 2222222 )(4)(4)( ycxaycxaycx 2222222 )(4242 ycxaxccxaxccx 22 )( ycxa a xc or © iTutor. 2000-2013. All Rights Reserved
  7. 7. By Squaring i.e. , Hence any point on the hyperbola satisfies 22 2 )( ycxa a xc 2222 2 22 22 yxccxxca a cx or 222 2 222 )( acy a acx or 1 )( )( 22 2 222 222 ac y aca acx or 12 2 2 2 b y a x (Since b2 = c2 – a2) 12 2 2 2 b y a x © iTutor. 2000-2013. All Rights Reserved
  8. 8. Conversely, let P( x, y) satisfy the above equation with 0 < a< c. Then, Therefore, Similarly, 2 22 22 a ax by 22 )( ycxPF 2 22 22 )( a ax bcx 2 22 222 )()( a ax accx (Since b2 = c2 – a2)x a c aPF x c a aPG © iTutor. 2000-2013. All Rights Reserved
  9. 9. In hyperbola c> a; and since P is to the right of the line x= a, x> a, Therefore, becomes negative. Thus, Therefore, Also, note that if P is to the left of the line x = –a, then In that case PF – PG = 2 a So, any point that satisfies lies on hyperbola Thus, we proved that the equation of hyperbola with origin (0,0) and transverse axis along x-axis is .ax a c x c a a ax a c PG aax a c x a c aPGPF 2 ,x a c aPF x c a aPG 12 2 2 2 b y a x 12 2 2 2 b y a x © iTutor. 2000-2013. All Rights Reserved
  10. 10. 222 2 2 2 2 where,1 acb b y a x The equation for a hyperbola can be derived by using the definition and the distance formula. The resulting equation is: aa cb b This looks similar to the ellipse equation but notice the sign difference. To graph a hyperbola, make a rectangle that measures 2a by 2b as a sketching aid and draw the diagonals. These are the asymptotes. © iTutor. 2000-2013. All Rights Reserved
  11. 11. Standard equation of a hyperbola with its center at the origin and vertical transverse axis For a hyperbola with its center at the origin and has the transverse axis horizontal, the standard equation is: c2 = a2 + b2 2 2 2 2 1 y x a b The equations of its asymptotes are: x a y b Focus: (0, c) (0, -c) (0, c) vertices: (0, a)(0, -a) (0, a) (0, -c) (0, c) (0, -a) (0, a) © iTutor. 2000-2013. All Rights Reserved
  12. 12. The center of the hyperbola may be transformed from the origin. The equation would then be: horizontal transverse axis 12 2 2 2 b ky a hx 12 2 2 2 b hx a ky vertical transverse axis The axis is determined by the first term NOT by which denominator is the largest. If the x term is positive it will be horizontal, if the y term is the positive term it will be vertical. © iTutor. 2000-2013. All Rights Reserved
  13. 13. Eccentricity The eccentricity (usually shown as the letter e), it shows how "uncurvy" (varying from being a circle) the hyperbola is. On this diagram:  P is a point on the curve,  F is the focus and  N is the point on the directrix so that PN is perpendicular to the directrix. The ratio PF/PN is the eccentricity of the hyperbola (for a hyperbola the eccentricity is always greater than 1). It can also given by the formula: F PN Directrix a ba e 22 © iTutor. 2000-2013. All Rights Reserved
  14. 14. Latus rectum of Hyperbola Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola. Directrix F Latus rectum The length of the Latus Rectum is a b2 2 © iTutor. 2000-2013. All Rights Reserved
  15. 15. The End Call us for more Information: www.iTutor.com Visit 1-855-694-8886

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