Introduction to learn hyperbolic geometry:

 In mathematics, hyperbolic geometry is one of important topics in geometry. Hyperbolic geometry is also called non-Euclidean geometry. Euclidean geometry is replaced by parallel postulate. In Euclidean geometry, parallel postulate is equivalent to the statement. The statement is in two dimensional spaces, for the given line l and point P not on l, when l is not intersect then P passes through exactly one line which is parallel to the line l. When the parallel postulate is false in hyperbolic geometry, two distinct lines through P do not intersect the line l.

  

Important concepts for learning Hyperbolic geometry:

 

Some important concepts to learn Hyperbolic geometry are as follows:

Focus :

    The fixed point is called a focus F1 (ae, 0) of the hyperbola.

Directrix :

        The fixed line is called the directrix of the hyperbola and its equation are x =a/e ..

Centre:

        The point of intersection of the transverse and conjugate axes of the hyperbola is called the centre of the hyperbola. Here C(0, 0) is called the centre of the hyperbola.

Vertices :

          The points of intersection of the hyperbola and its transverse axis are called its vertices. The vertices of the hyperbola are A(a, 0) and A′(− a, 0).As in the case of ellipse, hyperbola also has the special property of the second focus F2(− ae, 0) and the second directrix x = − a/e .

Standard equation of the hyperbola:

        x2 / a2y2 / b2 = 1 where b2 = a2 (e2 − 1) is a positive quantity. This is the required standard equation of the hyperbola. Eccentricity e, (e > 1) for hyperbola

 

Solved problems for learning hyperbolic geometry:

 

Some solved problems for learning hyperbolic geometry are as follows:

Example 1:

          Find the equation of the hyperbola whose transverse axis is along x-axis. The centre is (0, 0) length of semi-transverse axis is 6 and eccentricity is 5.

Solution:

   Since the transverse axis is along x-axis and the centre is (0, 0), the equation of the hyperbola is of the form
                         x2/a2 − y2/b2 = 1

   Given that semi-transverse axis a = 6, eccentricity e = 5

 We know that b2 = a2 (e2 − 1)

                     b2 = 36(24)= 864

The equation of the hyperbola is x2/36 − y2/864 = 1

 

Example 2:

    Find the equation of the hyperbola whose transverse axis is parallel to y-axis, centre (0, 0), and length of semi-conjugate axis is 4 and eccentricity is 3.

Solution:

  From the given data the hyperbola is of the form  y2/a2 −x2/b2 = 1

  Given that semi-conjugate axis b = 4 and e = 3,

                                     b2 = a2 (e2 − 1)

                                    42 = a2 (32 − 1)

                                    a2 = 16 / 8 = 2

  Hence the equation of the hyperbola is y2 / 2 −  x2/ 16 = 1