# Transverse and Conjugate Axis of the Hyperbola Definition, Examples | How to find Length of Transverse and Conjugate Axes of Hyperbola?

The transverse and conjugate axis of the hyperbola is here. Check the definition of the transverse and conjugate axis. Refer to Standard forms of the hyperbola with the center, vertices, and foci. Learn How to find the Length of Transverse and Conjugate Axes of a Hyperbola. Know the procedure to solve hyperbola problems along with solutions.

### Hyperbola – Definition

The hyperbola is a set of all the points in such a way that the difference of distance between any of the points on the hyperbola to the fixed points is always constant. The fixed points of the hyperbola are called the “foci of hyperbola”. The hyperbola graph is not continuous i.e., every hyperbola has two distinct points or branches. The transverse axis is nothing but the line segment where both endpoints of the hyperbola are on it.

The transverse axis endpoints are known as vertices of the hyperbola. “Center” is the point halfway between the foci which is the midpoint of the traverse axis. The transverse axis of hyperbola x2 / a2 – y2 / b2 = 1 is along the x-axis and the length of a hyperbola is 2a.

### Important Formulae and Terms of Hyperbola

There are a few terms related to hyperbola which has to be understood to get perfection in this concept. The important terms used in hyperbola are:

• Eccentricity: 1 + [(transverse axis)2 + (conjugate axis)2]
• Directrix: x = (-a/e), x = (a/e)
• Focii: S’ = (-ae,0), S = (ae,0)
• Conjugate Axis: The line segment of length 2b, between 2 points B’ = (0,-b) & B = (0,b) is called hyperbola conjugate axis.
• Transverse Axis: The line segment of the length 2a in which focii ” S’ ” and “S” lie is called the hyperbola transverse axis.
• Principle Axis: The conjugate and transverse axis both combinedly called a principle axis.
• Vertices: A’ = (-a, 0) & A = (a, 0)
• Double Ordinate: The chord that is perpendicular to the transverse axis is known as double ordinate.
• Focal Chord: A chord that passes through a focus is known as a focal chord.
• Latus Rectum: The focal chord which is perpendicular to the transverse axis is called the latus rectum.

The length of latus rectum = [(conjugate)2 / transverse] = (2b2 / a) = 2a (e2 – 1)

The difference of the focal distances is the constant value

i.e., |PS-PS’| = 2a

Length of latus rectum = 2e * (the distance of the focus from the corresponding directrix)

Endpoints of Latus Rectum: (± ae, ± b2 / a)

Centre: The point at which bisects every chord of the conic which is drawn through it and is called the center of the conic.

C: (0, 0) is the centre of [(x2/a2) – (y2/b2)] = 1

### Key Points about Transverse and Conjugate Axis of the Hyperbola

• Any point present on the conjugate hyperbola will be in the form (a tan θ, b sec θ).
• The equation of the hyperbola conjugate to xy = c2 is xy = -c2
• Conjugate Hyperbola + Hyperbola = 2 (Pair of Asymptotes).
• The point where a pair of the diameter of conjugates meets a hyperbola and the form of the conjugates of a parallelogram, where vertices lie on asymptotes and the area is constant.
• The equation of asymptotes and hyperbola differ by the similar constant by which the asymptotes equations and conjugate hyperbola differ.
• If a pair of diameters of hyperbola conjugate meet the hyperbola and its conjugate in T, T’ and R, R’ respectively. Then the asymptotes bisect TR and TR’.
• If e1 and e2 are the hyperbola eccentricities and its conjugate then the equation is e1-2 + e2-2 = 1
•  Two hyperbolas with similar eccentricity are said to be similar.
• The focus of conjugate and hyperbola are concyclic and they form the square vertices.

### Standard Forms of Hyperbola Equation with Center (0,0)

The standard form of hyperbola equation with center (0,0) and the transverse axis on x-axis is x2 / a2 – y2 / b2 = 1

where,

• the transverse axis length is 2a
• the vertices coordinates are (±a,0)
• the conjugate axis length is 2b
• the co-vertices coordinates are (0, ±b)
• the distance between foci is 2c, where c2=a2 + b2
• the foci coordinates are (±c,0)
• the asymptotes equation is y = ±b/a x

The standard form of hyperbola equation with center (0,0) and the transverse axis on y-axis is y2 / a2 – x2 / b2 = 1

where,

• the transverse axis length is 2a
• the vertices coordinates are (0,±a)
• the conjugate axis length is 2b
• the co-vertices coordinates are (±b,0)
• the distance between foci is 2c, where c2=a2 + b2
• the foci coordinates are (0,±c)
• the asymptotes equation is y = ±a/b x

### Standard Forms of Hyperbola Equation with Vertices and Foci

• First of all, determine if the transverse axis lies on the x-axis or y-axis. Check that a2 is under the variable with +(positive) coefficient. Therefore, if you set another variable equal to zero (0), you can find the intercepts easily. The point at which the intercepts of hyperbola coincide with vertices, it is centered as an origin.
• If the equation is of the form x2 / a2 – y2 / b2 = 1, then the axis of the transverse line lies on the x-axis. The vertices are pointed at (±a,0) and foci are pointed at (±c,0).
• If the equation is of the form y2 / a2 – x2 / b2 = 1, then the axis of the transverse line lies on the y-axis. The vertices are pointed at (0,±a) and the foci are pointed at (0,±c)
• Solve the value for a, by using the equation a = √a2
• Solve the value for c, using the equation c = √a2 + b2

### Transverse and Conjugate Axes of Hyperbola Examples

Problem 1:

Find the lengths of conjugate and transverse axis of the hyperbola 16x2 – 9y2 = 144?

Solution:

The given equation of the hyperbola is 16x2 – 9y2 = 144

x2/9 – y2/16 = 1 is the (1) equation

The above equation (1) is of the form x2 / a2 – y2 / b2 = 1, where a2 = 9 and b2=16

Therefore, the length of the transverse axis is 2a, which can be written as 2*3 = 6 and the length of the conjugate axis is 2b, which can be written as 2*4 = 8

Problem 2:

Find the lengths of conjugate and transverse axis of the hyperbola 3x2 – 6y2 = -18?

Solution:

The given equation of the hyperbola is 3x2 – 6y2 = -18

The hyperbola equation can be written as x2/6 – y2/3 = 1 is the (1) equation

The above equation (1) is of the form x2 / a2 – y2 / b2 = -1, where a2 = 6 and b2=3

Hence, the length of the transverse axis is equal to 2b which is 2 * √3 = 2√3 and the length of the conjugate axis is equal to 2a which is 2 * √6 = 2√6

Problem 3:

Find the hyperbola eccentricity whose latus rectum is half of the transverse axis?

Solution:

Let the equation of the hyperbola be x2 / a2 – y2 / b2 = 1

Transverse axis = 2a and lactus-rectum = (2b2/a)

According to the question, (2b2/a) = (1/2) * 2a

2b2=a (Since, b2=a2(e2-1))

2a2(e2-1) = a2

2e2-2 = 1

e2=(3/2)

e = √(3/2)

Therefore, hyperbola eccentricity is √(3/2).

Thus, the final solution is √(3/2).