The eccentricity of the conjugate hyperbola to the hyperbola $$x^2-3y^2$$ = 1 is

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, $$b^2$$ = $$1\over 3$$

$$\therefore$$  eccentricity e = $$\sqrt{1 + a^2/b^2}$$ = $$\sqrt{1+3}$$ = 2

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