For the hyperbola x2100-y225-1-The eccentricity is 5

The correct option is B FalseFor hyperbola x^2a^2 y^2b^2=1 e=√(a^2+b^2b^2) For hyperbola x^210^2 y^25^2=1 e=√(10^2+5^25^2) √(12525)=√(5) ...

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Standard XI

Mathematics

Standard Form of an Ellipse

For the hyper...

Question

For the hyperbola $\frac{x^{2}}{100} - \frac{y^{2}}{25} = 1$ ,The eccentricity is $5$

True

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False

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Solution

The correct option is B False
For hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 e = \sqrt{\frac{a^{2} + b^{2}}{b^{2}}}$
For hyperbola $\frac{x^{2}}{10^{2}} - \frac{y^{2}}{5^{2}} = 1 e = \sqrt{\frac{10^{2} + 5^{2}}{5^{2}}} \sqrt{\frac{125}{25}} = \sqrt{5}$

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Similar questions

Q. For the hyperbola

\frac{x^{2}}{100} - \frac{y^{2}}{25} = 1

, prove that eccentricity

= \frac{\sqrt{5}}{2}

Q. If the hyperbola

\frac{x^{2}}{81} - \frac{y^{2}}{25} = 1

, then find the value of eccentricity

$I f t h e e c c e n t r i c i t y o f h y p e r b o l a x^{2} - y^{2} s e c^{2} α = 5 i s \sqrt{3} t i m e s t h e e c c e n t r i c i t y o f t h e e l l i p s e x^{2} s e c^{2} α + y^{2} = 25, t h e n α =$

The eccentricity of the hyperbola $x^{2} - y^{2} = 25$ is

Q. If the eccentricity of the hyperbola

x^{2} - y^{2} {sec}^{2} α = 5

\sqrt{3}

times the eccentricity of the ellipse

x^{2} {sec}^{2} α + y^{2} = 25

, then a value of

α

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For the hyperbola x2100−y225=1,The eccentricity is 5

The correct option is B FalseFor hyperbola x2a2−y2b2=1e=√a2+b2b2For hyperbola x2102−y252=1e=√102+5252√12525=√5

For the hyperbola $\frac{x^{2}}{100} - \frac{y^{2}}{25} = 1$ ,The eccentricity is $5$

The correct option is B False
For hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 e = \sqrt{\frac{a^{2} + b^{2}}{b^{2}}}$
For hyperbola $\frac{x^{2}}{10^{2}} - \frac{y^{2}}{5^{2}} = 1 e = \sqrt{\frac{10^{2} + 5^{2}}{5^{2}}} \sqrt{\frac{125}{25}} = \sqrt{5}$