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 tanθ, b secθ)
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 (Pair of 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)/b²
Parametric Coordinates	(a sec θ, b tan θ)	(a tan θ, b sec θ)
Length of Latus Rectum	2b²/a	2a²/b
Tangent at 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 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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