# 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.

(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!