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.

Some important concepts to learn Hyperbolic geometry are as follows:

**Focus :**

The fixed point is called a focus *F*_{1} (*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 *F*2(− *ae*, 0) and the second
directrix *x* = − *a*/*e* .

**Standard equation of the hyperbola:**

** x^{2} / a^{2} − y^{2} / b^{2} = 1** where

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

x^{2}/a^{2} −
y^{2}/b^{2} = 1

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

We know that b^{2} = a^{2} (e^{2} − 1)

b^{2} = 36(24)= 864

The equation of the hyperbola is **x ^{2}/36 − y^{2}/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 y^{2}/a^{2} −x^{2}/b^{2} = 1

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

b^{2} = a^{2} (e^{2} − 1)

4^{2} = a2 (3^{2} − 1)

a^{2} = 16 / 8 = 2

** Hence the equation of the hyperbola is y ^{2} / 2 − x^{2}/ 16 = 1**