### 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.

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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.

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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)?

Answer:

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.

Answer:

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.