Equation of Conjugate Hyperbola

Here you will learn what is conjugate hyperbola, equation of conjugate hyperbola and basic definitions of like eccentricity and latus rectum.

Let’s begin –

What is Conjugate Hyperbola ?

The hyperbola whose transverse & conjugate axes are respectively the conjugate and transverse axes of given hyperbola is called the conjugate hyperbola of given hyperbola. 

The equation of the conjugate hyperbola of the hyperbola \(x^2\over a^2\) – \(y^2\over b^2\) = 1 is

-\(x^2\over a^2\) + \(y^2\over b^2\) = 1.

The graph of the conjugate hyperbola is shown in figure.

The eccentricity of the conjugate hyperbola is given by \(a^2\) = \(b^2(e^2-1)\) and the length of latus rectum is \(2a^2\over b\).

Conjugate Hyperbola & Basic Definitions :

The equation of the conjugate hyperbola is -\(x^2\over a^2\) + \(y^2\over b^2\) = 1.conjugate hyperbola

(a) Centre (0,0).

(b) Vertices (0,b) & (0,-b)

(c) foci \((0, \pm ae)\)

(d) Length of  transverse axis is 2b

(e) Length of  conjugate axis is 2a

(f) Equation of  directrices is x = \(\pm {b\over e}\)

(g) Eccentricity is given by \(a^2\) = \(b^2(e^2-1)\)

(h) Length of latus rectum is \(2a^2\over b\).

(i) Equation of the transverse axis is x = 0.

(j) Equation of the conjugate axis is y = 0.

Example : Find the eccentricity of the conjugate hyperbola to the hyperbola \(x^2 – 3y^2\) = 1.

Solution : Equation of the conjugate hyperbola to the hyperbola \(x^2 – 3y^2\) = 1 is \(-x^2 + 3y^2\) = 1 \(\implies\) -\(x^2\over 1\) + \(y^2\over 1/3\) = 1.

Here \(a^2\) = 1, \(b^2\) = 1/3

Therefore eccentricity is e = \(\sqrt{1 + a^2/b^2}\) = 2.

Hope you learnt conjugate hyperbola, learn more concepts of hyperbola and practice more questions to get ahead in the competition. Good luck!

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