Hyperbola is a conic section which has two symmetrical open curves. It is formed by the intersection between a plane and two cones which lies on opposite sides of the same vertex with its pointed top facing each other.

Hyperbola can be defined as a set of points lying in a plane such that the ratio it forms of distance from fixed point foci, either one of the focus to its distance from a line called the directrix is a constant denoted by the letter ‘$e$’. The letter ‘$e$’ is known as the eccentricity and carries a value greater than 1 in case of hyperbola

It can also be defined as locus of points P such that the difference of any point P from either of its foci F1 or F2 is always constant denoted by the letter ‘$k$’.
There are two conditions to draw the hyperbola, one is when the transverse axis is horizontal which mean the center, foci and vertices all three lies on the same line which is parallel to the $x$ axis. The other condition is when the transverse axis is vertical that is when all three center, focus and vertices lies on the same line which is parallel to the $y$ axis. The eccentricity of hyperbola is always greater than $1$ (Eccentricity is the measure of how much is the conic figure curved, $e$ = $\frac{c}{a}$)

• Horizontal Transverse Axis 

Horizontal Transverse Axis

$\frac{(x - h)^{2}}{a^{2}}$ - $\frac{(y - k)^{2}}{b^{2}}$ = $1$

Center lying at $(h,\ k)$, $a$ is the distance between the center and the vertex, $c$ is the distance between the center and the foci, $b$ is half the height of the rectangular box. Equation of asymptote is $y$ = $\pm$ $(\frac{b}{a})$ $ \times\ (x - h)\ +\ k$
• Vertical Transverse Axis

Vertical Transverse Axis

$\frac{(y - k)^{2}}{a^{2}}$ - $\frac{(x - h)^{2}}{b^{2}}$ = $1$

Center lying at $(h,\ k)$, a is the distance between the center and the vertex, $c$ is the distance between the center and the foci, $b$ is the half the width of the rectangular box. Equation of asymptote is $y$ = $\pm$ $(\frac{a}{b})$ $\times\ (x - h)\ +\ k$
Eccentricity is defined as the ratio of the distance from any point lying n the hyperbola to the foci to the distance from the same point lying on the hyperbola to the fixed straight line called directrix. This ratio always remains fixed with respect to a hyperbola and has a value greater than one.

Eccentricity of hyperbola is denoted by the formula

$'e$ = $\sqrt{1\ +\ \frac{b^{2}}{a^{2}}}$

Where, $a$ = length of semi major axis, $b$ = length of semi minor axis

Alternate formula of eccentricity of hyperbola is:

$‘e$ = $\frac{c}{a}$

Where, $c$ = distance from the center to either of the foci, $a$ = length of semi major axis
Latus Rectum is defined as a chord of the hyperbola. It passes through either of the two foci and is perpendicular to the transverse axis. Also, parallel to the fixed straight line called directrix. A hyperbola has a pair of latus rectum

Formulas of latus rectum are as follows:

Coordinates of the end points of the latus rectum of hyperbola $(ae,$ $\frac{-b^{2}}{a})$ and $(ae,$ $\frac{b^{2}}{a})$. Hence, the length of the latus rectum is the distance between the end points of the latus rectum $(ae\ –\ ae,$ $\frac{b^{2}}{a}$$(\frac{-b^{2}}{a}))$ = $2$ $\frac{b^{2}}{a}$. Alternate formula of the length of latus rectum of hyperbola is $2a\ (e^{2}\ –\ 1)$ [where, $a$ is the length of the major axis, $b$ is the length of the minor axis, $e$ = eccentricity]
Transverse axis of a hyperbola is defined as the line starting from one of the vertex passing through the foci and the center and ending on another vertex of the hyperbola. It is along the $x$ axis and the length of it is $2a$. Conjugate axis is defined as the line which is perpendicular to the transverse axis and passes through the center of the hyperbola. The length of the conjugate axis is given by $2b$. The condition when in a hyperbola the transverse axis is equal to the conjugate axis, then it is called a rectangular hyperbola.

Therefore, length of transverse axis = length of conjugate axis

$2a$ = $2b$ or $a$ = $b$

Derivation of equation of rectangular hyperbola,

Standard equation of hyperbola is given by $\frac{x^{2}}{a^{2}}$$\frac{y^{2}}{b^{2}}$ = $1$

Substituting the condition of rectangular hyperbola b = a in the standard equation, we get

$\frac{X^{2}}{a^{2}}$$\frac{y^{2}}{a^{2}}$ = $1$

$X^{2}\ –\ y^{2}$ = $a^{2}$ [Hence derived equation of rectangular hyperbola]

Alternate definition of rectangular hyperbola could be, when the asymptotes of a hyperbola are perpendicular to each other or making right angle with each other. It is also known as equilateral hyperbola or right hyperbola. The eccentricity of rectangular hyperbola is rad2 and the angle between the asymptote is $tan^{-1}$ $(\frac{b}{a})$ = $\frac{\pi}{4}$ or, $\frac{b}{a}$ = tan $(\frac{\pi}{4})$ or $\frac{b}{a}$ = $1$ or $b$ = $a$ which shows length of conjugate axis must be equal to the length of transverse axis.
Example 1:

Find the center, vertices, axis, foci, eccentricity and asymptotes of the hyperbola $x^{2}\ –\ 4y^{2}\ +\ 2x\ –\ 16y\ –\ 19$ = $0$

Solution: 

Arrange the given problem in the standard form $\frac{(x – h)^{2}}{a^{2}}$$\frac{(y – k)^{2}}{b^{2}}$ = $1$

To get that we first take the loose number -$19$ on the other side of the equal to $x^{2}\ –\ 4y^{2}\ +\ 2x\ –\ 16y$ = $19$

Arrange the $x$ terms and $y$ terms together $x^{2}\ +\ 2x\ –\ 4y^{2}\ –\ 16y$ = $19$

Factor the coefficient of the $x^{2}$ from the $x$ terms and similarly factor out the coefficient of the $y^{2}$ from the $y$ terms $1(x^{2}\ +\ 2x)\ –\ 4(y^{2}\ +\ 4y)$ = $19$

Complete the square within the parenthesis $1(x^{2}\ +\ 2x + 1)\ -\ 4(y^{2}\ +\ 4y + 4)$ = $19$  = $19\ +\ 1\ –\ 16$
or $ (x + 1)^{2}\ –\ 4(y + 2)^{2}$ = $4$

Divide throughout by the loose number $4$ present on the right side of the equal to

$\frac{(x + 1)^{2}}{4}$$\frac{4(y + 2)^{2}}{4}$ = $\frac{4}{4}$

$\frac{(x + 1)^{2}}{2^{2}}$$\frac{(y + 2)^{2}}{1^{2}}$ = $1$

Center : $(-1,\ -2)$

Vertices :$(h\ +\ a,\ k)$ and $(h\ –\ a,\ k)$ = $(-1\ +2,\ -2)$ and $(-1-2,\ -2)$ = $(1, -2),$ $(-3, -2)$

Axes : $a$ = $2$, $b$ = $1$

Foci :$(h + c, k)$ and $(h – c, k)\ where \ c$ = $\sqrt{(a^{2}\ +\ b^{2})}$ = $\sqrt{5}$. 
Therefore foci = $(-1\ +\ \sqrt{5},\ -2)$ and $(-1\ - \sqrt{5},\ -2)$

Eccentricity = $\frac{c}{a}$ = $\frac{\sqrt{5}}{2}$

Asymptotes $y$ = $\pm$ $\frac{b}{a}$ $(x - h)\ +\ k$
or $y$ = $\pm$$\frac{1}{2}$ $(x\ +\ 1)$ - $2$