HYPERBOLA

Created: Monday, 12 September 2011 10:01 | Published: Monday, 12 September 2011 10:01 | Written by Super User | Print | Email

Introduction to Hyperbolas

A hyperbola is an open curve with two branches, the intersection of a plane with both halves of a double cone.

The two fixed points F1 and F2 are called the foci of the hyperbola. The midpoint of the line segment joining the foci is called the centre of the hyperbola. The line through the foci is called the transverse axis and the line through the centre and perpendicular to the transverse axis is called the conjugate axis. The points at which the hyperbola intersects the transverse axis are called the vertices of the hyperbola.

In the above figure, let the distance between the two foci F₁ and F₂ be 2c, and the distance between the two vertices A and B be 2a. With the assumed information, we can define b as

b = √c² – a²

Also 2b is the length of the conjugate axis.

By taking the point P at A and B in the above figure, we have

BF₁ – BF₂ = AF₂ – AF₁ (By definition)

BA +AF₁– BF₂ = AB + BF₂– AF₁

i.e., AF₁ = BF₂

So that, BF₁ – BF₂ = BA + AF₁– BF₂ = BA = 2a

Equation of Hyperbola

Above diagrams represent two possible orientations of hyperbola.

Now, we will derive the equation for the hyperbola with foci on the x - axis. Consider the following diagram:

Let F₁ and F₂ be the foci and O be the mid-point of the line segment F₁F₂. Let O be the origin and the line through O via F₂ be the positive x – axis and that through F₁ as the negative x – axis. Let the coordinates of F₁ be (-c, 0) and F₂ be (c, 0).

Let P(x, y) be any point on the hyperbola such that the difference between the farther and closer point be 2a

∴ PF₁ – PF₂ = 2a - (i)

Applying distance formula in the above equation:
√(x + c) ² + y² - √(x - c) ² + y² = 2a

√(x + c) ² + y² = 2a + √(x - c) ² + y²

On squaring both sides, we get:
(x + c) ² + y² = 4a² + 4a√(x - c) ² + y² + (x – c) ² + y²

On further simplifying:
cx/a – a = √(x - c) ² + y²

On squaring both sides again and simplifying further:
x²/a² – y²/(c² – a²) = 1

x²/a² – y²/b² = 1 (Since c² – a² = b²)

Hence, any point on the hyperbola satisfies

x²/a² – y²/b² = 1

To summarise, here are the observations from the standard equations of parabola:

1. Hyperbola is symmetric with respect to both the axes, since if (x, y) is a point on the hyperbola, then (– x, y), (x, – y) and (– x, – y) are also points on the hyperbola.

2. The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.

Eccentricity

For the hyperbola, x²/a² – y²/b² = 1

We have, b² = a² (e² – 1)

e² = (a² + b²)/a²

e² = 1 + b²/a²

e = √1 + b²/a²

e = √1 + (2b)²/(2a)²

e = √1 + (Conjugate axis)²/(Transverse axis)²

Latus Rectum

Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

Length of latus rectum in hyperbola is 2b²/a

Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor!

About eAge Tutoring:

eAgeTutor.com is the premium online tutoring provider. Using materials developed by highly qualified educators and leading content developers, a team of top-notch software experts, and a group of passionate educators, eAgeTutor works to ensure the success and satisfaction of all of its students.

Contact us today to learn more about our tutoring programs and discuss how we can help make the dreams of the student in your life come true!

Reference Links:

Category: ROOT

Articles