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The eccentricity of the hyperbola whose latus-rectum is 8 and conjugate axis is equal to half the distance between the foci is
- a)4/3
- b)4/√3
- c)2/√3
- d)none
Correct answer is option 'C'. Can you explain this answer?
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The eccentricity of the hyperbola whose latus-rectum is 8 and conjugat...
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The eccentricity of the hyperbola whose latus-rectum is 8 and conjugat...
The latus rectum of a hyperbola is given by the formula $LR = \frac{2b^2}{a}$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively.
In this problem, we are given that $LR = 8$. We are also given that the conjugate axis (which is twice the length of the semi-minor axis) is equal to half the distance between the foci. Let the distance between the foci be $2c$. Then, the conjugate axis is $\frac{2c}{2} = c$.
Since the eccentricity of a hyperbola is defined as $e = \frac{c}{a}$, we have $e = \frac{c}{a} = \frac{c}{\frac{1}{2}LR} = \frac{c}{\frac{1}{2}(8)} = \frac{c}{4}$.
Therefore, the eccentricity is $\boxed{\frac{1}{4}}$.
In this problem, we are given that $LR = 8$. We are also given that the conjugate axis (which is twice the length of the semi-minor axis) is equal to half the distance between the foci. Let the distance between the foci be $2c$. Then, the conjugate axis is $\frac{2c}{2} = c$.
Since the eccentricity of a hyperbola is defined as $e = \frac{c}{a}$, we have $e = \frac{c}{a} = \frac{c}{\frac{1}{2}LR} = \frac{c}{\frac{1}{2}(8)} = \frac{c}{4}$.
Therefore, the eccentricity is $\boxed{\frac{1}{4}}$.
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The eccentricity of the hyperbola whose latus-rectum is 8 and conjugate axis is equal to half the distance between the foci isa)4/3b)4/√3c)2/√3d)noneCorrect answer is option 'C'. Can you explain this answer?
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