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: 

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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