If the eccentricity of the hyperbola x2-a2 - y2-b2 - 1 is e then the eccentricity of the hyperbola y2-b2 - x2-a2 - 1 is -

The correct option is D e/√(e^2 1) For hyperbola x^2/a^2 y^2/b^2 = 1 eccentricity ⇒ e=√(1+b^2a^2)⇒b^2a^2=e^2 1 For hyperbola x^2/b^2 ...

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Standard XII

Physics

Multiplication with Vectors

If the eccent...

Question

If the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $e$ then the eccentricity of the hyperbola $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ is :

e

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\frac{e}{\sqrt{e^{2} - 1}}

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e \sqrt{e^{2} - 1}

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e^{2} - e

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Solution

The correct option is D $\frac{e}{\sqrt{e^{2} - 1}}$
For hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
eccentricity $\Rightarrow e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \Rightarrow \frac{b^{2}}{a^{2}} = e^{2} - 1$
For hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{a^{2}} = 1$
Required eccentricity $e^{'} = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{1}{e^{2} - 1}} = \frac{e}{\sqrt{e^{2} - 1}}$

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Similar questions

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\int_{1}^{3} f f f . . . . . f (e)      n t i m e s d e

is equal to

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If the eccentricity of the hyperbola x2a2−y2b2=1 is e then the eccentricity of the hyperbola y2b2−x2a2=1 is :

The correct option is D e√e2−1For hyperbola x2a2−y2b2=1 eccentricity ⇒e=√1+b2a2⇒b2a2=e2−1For hyperbola x2b2−y2a2=1 Required eccentricity e′=√1+a2b2=√1+1e2−1=e√e2−1

If the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $e$ then the eccentricity of the hyperbola $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ is :