# Hyperbola

Hyperbola is a subdivision of conic sections in the field of mathematics. When the surface of a cone intersects with a plane, curves are formed, and these curves are known as conic sections.

There are three categories of conic sections: the eclipse, the Hyperbola, and the parabola. We use conic sections to study 3D geometry that has a vast number of applications in various fields of engineering. Also, when spacecraft uses the gravitational slingshot technique then the path followed by the craft is a hyperbola.  Before looking at the eccentricity of hyperbola formula let us try to understand the meaning of Hyperbola and Hyperbola definition first.

### Define Hyperbola

According to the Hyperbola definition, it is a collection of points in the plane such that there is a constant distance between two fixed points and each point. Hyperbola is made up of two similar curves that resemble a parabola.

Let us consider the Hyperbola in the above diagram. The Two fixed points $F_{1}$ and $F_{2}$ in the diagram are known as foci or focus. Now consider three points on the hyperbola $P_{1}$, $P_{2}$, and $P_{3}$. According to the definition of a hyperbola, we determine that $P_{1}$$F_{1}$ + $P_{1}$$F_{2}$ is equal to constant. Similarly, $P_{2}$$F_{1}$ + $P_{2}$$F_{2}$ and $P_{3}$$F_{1}$+$P_{3}$$F_{2}$ are also constant.

$P_{1}$$F_{1}$+$P_{1}$$F_{2}$=$P_{2}$$F_{1}$+$P_{2}$$F_{2}$=$P_{3}$$F_{1}$+$P_{3}$$F_{2}$= Constant

So, when we join both the foci using a line segment, then its midpoint gives us centre (O). Hence, this line segment is known as the transverse axis.

### Eccentricity of Hyperbola

The eccentricity of Hyperbola is the distance ratio from the centre to a vertex and from the centre to a focus(foci). The eccentricity of Hyperbola formula can be showed as follows:

Eccentricity(e)=$\frac{c}{a}$

Eccentricity is never less than 1 for Hyperbola. Since c ≥ a.

### Standard Equation of Hyperbola

Let us now derive the standard equation of Hyperbola. For this, consider a hyperbola with centre (O) at (0,0) and its foci lie on any one of the x or y-axis. Look at the diagram below to understand the concept thoroughly.

Both the foci's lie at a distance of "c" on the x-axis and the vertices are at a distance "a" from (0,0) origin. Let us consider a point Z on the Hyperbola so that it satisfies the definition Z$F_{1}$+Z$F_{2}$ is constant 2a.

2a = Z$F_{1}$+Z$F_{2}$

According to distance formulae

$\sqrt{(x + c)^{2} + y^{2}}$ - $\sqrt{(x - c)^{2} + y^{2}}$ = 2a       --equation(1)

$\sqrt{(x + c)^{2} + y^{2}}$ = 2a + $\sqrt{(x - c)^{2} + y^{2}}$

Squaring both sides

$\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{c^{2}}$ - $a^{2}$ = 1

Since we know that $b^{2}$ = $c^{2}$ - $a^{2}$ and 0 < a < c.

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

Substituting $y^{2}$ in equation(1)

$\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1

### Solved Examples

Example 1:

Given the hyperbola $\frac{x^{2}}{9}$ - $\frac{y^{2}}{25}$ = 1 Then, what is the position of Point P(6,-5)?

Hyperbola definition fits perfectly for $\frac{x^{2}}{9}$ - $\frac{y^{2}}{25}$ = 1

According to the standard equation of Hyperbola

$\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ = 1

Rearranging this equation, we know that P lies either inside or outside of the Hyperbola.

Hence,

$\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ - 1 < 0, or > 0

Substitute the values of x, y, a, and b in this equation.

$\frac{6^{2}}{9}$ - $\frac{(-5)^{2}}{25}$ - 1

Further solving this equation

= $\frac{36}{9}$ - $\frac{25}{25}$ - 1

= 4 - 1 - 1

= 2

Since 2>0, we can conclude that the point P lies inside the given Hyperbola.

Example 2:

Find the equation of Hyperbola whose vertices are (9,2) and (1,2) as well as the distance between the foci is 10.

According to the meaning of Hyperbola the distance between foci of Hyperbola is 2ae

2ae=10

In the eccentricity of Hyperbola formula

ae=5 --(1)

Since both, the vertices are at two on the y-axis.

We can calculate the centre of the Hyperbola by finding the midpoint of vertices.

$\frac{(9+1)}{2}$, $\frac{(2+2)}{2}$

=(5,1)

Assume the equation of this hyperbola is $\frac{(x-p)^{2}}{a^{2}}$ - $\frac{(y-q)^{2}}{b^{2}}$ = 1

length of the transverse axis= 8

2a=8

a=4

Using (1)

4e=5

e=5/4

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

Substituting respective values

$b^{2}$ = 9

Now let’s assemble the equation

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

$\frac{(x-5)^{2}}{16}$ - $\frac{(y-2)^{2}}{9}$ = 1

Solving the equation further

$9x^{2} - 16y^{2} - 90x - 64y + 17$ = 0

### Did you know?

The design of cooling towers in Nuclear reactors are hyperbolic structures. They help make the structure durable, efficient, and cost-effective.