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Conjugate hyperbola

What do you mean by a conjugate hyperbola?

The hyperbola whose transverse and conjugate axes respectively are the conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola of the given hyperbola. The asymptotes of these two hyperbolas are also the same.

The conjugate hyperbola of the hyperbola x2/a2 – y2/b2 = 1 is x2/a2 – y2/b2 = -1.

 Its transverse and conjugate axes are along y and x axes respectively.

 Some key Points:

  • Any point on the conjugate hyperbola is of the form (a tan q, b sec q)

  • The equation of the conjugate hyperbola to xy = c2 is xy = –c2.

  • The equation of the hyperbola and asymptotes differ by the same constant by which the equations of the asymptotes and the conjugate hyperbola differ.

  • Hyperbola + Conjugate hyperbola = 2 (Asymptotes).

  • The tangents drawn at the points, where a pair of conjugate diameters meets a hyperbola and its conjugates form a parallelogram, whose vertices lie on the asymptotes and whose area is constant.

  • If a pair of conjugate diameters of hyperbola meet the hyperbola and its conjugate in P, P’ and D, D’ respectively, then the asymptotes bisect PD and PD’.

  • If e1 and e2 are the eccentricities of the hyperbola and its conjugate then e1-2 + e2-2 = 1.

  • Two hyperbolas with the same eccentricity are said to be similar.

  • The focus of a hyperbola and its conjugate are concyclic and and form the vertices of a square.

We now compare the hyperbola and the conjugate hyperbola closely by looking at their various parameters: 

 

Hyperbola

Conjugate Hyperbola

Standard Equation

x2/a2 – y2/b2 = 1

-x2/a2 + y2/b2 = 1

Centre

(0,0)

(0, 0)

Equation of Transverse axis

y = 0

x= 0

Equation of Conjugate axis

x = 0

y = 0

Length of Transverse axis

2a

2b

Length of Conjugate axis

2b

2a

Vertices

(±a, 0)

(0, ±b)

Foci

(±ae, 0)

(0, ±be)

Equation of directrix

x = ±a/e

y = ±b/e

Eccentricity

e = √(a2+b2)/a2

e = √(a2+b2)/a2

Parametric Coordinates

(a sec θ, b tan θ)

(b sec θ, a tan θ)

Length of Latus Rectum

2b2/a

2b2/a

Tangent to the vertices

x = ±a

y = ±b

 

 

 
Let us now study some of the illustrations based on conjugate hyperbola and the conjugate axis of a hyperbola:
Illustration:
If e1 and e2 are the eccentricities of the hyperbola x2/a2 – y2/b2 = 1 and its conjugate hyperbola, prove that e1–2 + e2–2 = 1.
Solution:

The eccentricity e1 of the given hyperbola is obtained from

b2 = a2(e12 – 1).                                                  …… (1)

The eccentricity of e2 of the conjugate hyperbola is given by

a2 = b2(e22 – 1).                                                  …… (2)

Multiply (1) and (2), we get,

1 = (e12 – 1)(e22 – 1) ⇒ 0 = e12 e22 – e12 – e22

⇒ e1–2 + e2–2 = 1.

Illustration:

Find the centre, eccentricity, foci, directrix and the lengths of the transverse and conjugate axes of the hyperbola whose equation is (x-1)2 – 2(y-2)2 + 6 = 0.

Solution:

The equation of the hyperbola can be written as (x-1)2 – 2(y-2)2 + 6 = 0.

We can rewrite it as Y2/(√3)2 – X2/(√6)2 = 1, where Y = (y-2) and X = (x-1)

Hence, we have the centre as X = 0, Y = 0 so (x-1) = 0and (y-2) = 0.

This gives x = 1 and y = 2.

This also means that a = √3 and b = √6 and hence the transverse axis = 2√3 and conjugate axis = 2√6.

We know b2 = a2(e2-1)

This means 6 = 3(e2-1) and so e = √3.

Hence, the foci are (1, 5) and (1, -1).

Equation of directrix is Y = ±a/e

Hence, directrices = y-2 = ± √3/√3 = ±1.

Hence, y = 3 or y = 1.                                                                                                                                                                      

Related Resources:

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