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  • If the eccentricities of the ellipse   <img alt="\mathrm{\frac{x^2}{4}+\frac{y^2}{3}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B4%7D&plu

If the eccentricities of the ellipse   \mathrm{\frac{x^2}{4}+\frac{y^2}{3}=1}  and the hyperbola

\mathrm{\frac{x^2}{64}-\frac{y^2}{b^2}=1}   are reciprocals of each other , then \mathrm{b^2} is

Option: 1

8


Option: 2

27


Option: 3

64


Option: 4

125


Answers (1)

Let  \mathrm{e \text { and } e^{\prime}}  be the eccentricities of hyperbola 

and its conjugate , respectively.

Given that,   \mathrm{2 \tan ^{-1}\left(\frac{b}{a}\right)=\frac{\pi}{3} \Rightarrow \frac{b}{a}=\frac{1}{\sqrt{3}}}

\mathrm{\begin{aligned} & \text { So, } e^2=1+\frac{1}{3}=\frac{4}{3} \\ & \frac{1}{e^{\prime 2}}+\frac{1}{e^2}=1 \\ & \Rightarrow \frac{1}{e^{\prime 2}}+\frac{3}{4}=1 \Rightarrow \frac{1}{e^{\prime 2}}=\frac{1}{4} \Rightarrow e^{\prime}=2 \end{aligned}}

Posted by

Ramraj Saini

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