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
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Any point on the conjugate hyperbola is of the form (a tanθ, b secθ)
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The equation of the conjugate hyperbola to xy = c2 is xy = –c2.
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The equation of the hyperbola and asymptotes differ by the same constant by which the equations of the asymptotes and the conjugate hyperbola differ.
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Hyperbola + Conjugate hyperbola = 2 (Pair of Asymptotes).
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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.
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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’.
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If e1 and e2 are the eccentricities of the hyperbola and its conjugate then e1-2 + e2-2 = 1.
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Two hyperbolas with the same eccentricity are said to be similar.
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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:
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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For getting an idea of the type of questions asked, refer the previous year papers.
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