**Conic sections: Standard equation and simple properties of Hyperbola**

The standard form of the equation of a hyperbola is developed in a similar methodology to an ellipse. Note, however, that a, b and c are related differently for hyperbolas than for ellipses.For a hyperbola, the distance between the foci and the centre is greater than the distance between the vertices and the centre.

This equation is the standard form of the equation of a hyperbola centred at the origin with the x-axis as its focal axis. When the y-axis is the focal axis:

The values *a*, *b*, and *c*, are associated by the distance formula *c*^{2} = *a*^{2} + *b*^{2}. Also, note that the length of the transverse axis = 2*a* and the length of the conjugate = 2*b.*

## Standard Form for the Equation of a Hyperbola (Center at (0,0))

As with circles and ellipses, hyperbolas do not have to be centred at the origin.

## Standard Form for the Equation of a Hyperbola (Center at (h,k))

Asymptotes are an essential aid for graphing a hyperbola. They help us determine its shape. The asymptotes mentioned are lines that the hyperbola approaches for large values of *x *and *y*.To find the asymptotes in the first case, we solve the equation for *y *to get:

As *x* enlarges, *a*^{2}/*x*^{2} goes closer to zero, therefore, as *x* → *∞*,* a*^{2}/*x*^{2} → 0. So, for large *x*,the value of *y =* ±(*b*/*a*)*x*. This shows that these lines are asymptotes of the hyperbola. The conjugate axis of a hyperbola is the line segment of length 2b joining (h, k + b) and (h, k – b) when the transverse axis is horizontal. When the transverse axis is vertical, the conjugate axis is the line segment of length 2b joining (h + b, k) and (h – b, k).

**Example 1. A hyperbola has the equation 9 x^{2} – 16y^{2} = 144. Find the vertices, foci, and asymptotes,**

and sketch the graph.

First, we divide both sides of the equation by 144 to put it into the standard form:

Since the *x*^{2}-term is +ive, horizontal transverse axis occurs.Its vertices and foci are on the *x*-axis. Since *a*^{2} = 16 and *b*^{2} = 9, we get *a =* 4, *b = *3, and $c=\sqrt{16+9}=5$.Thus, Vertices:(±4, 0), Foci: (±5, 0) and Asymptotes: y = ±¾x.

**Example 2. Find the equation and the foci of the hyperbola with vertices (0, ±2) and asymptotes y = ±2x.**

The hyperbola has a vertical transverse axis with a = 2 as the vertices are placed on the y-axis. From the asymptote equation, we see a/b = 2. We find a = 2, so 2/b = 2; thus, b = 1.

The equation is:

To find the foci,

*c*^{2} = *a*^{2} + *b*^{2}= 2^{2} + 1^{2} = 5 .So, *c* = $\sqrt{5}$

Thus, the foci are (0, ± $\sqrt{5}$).