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  • If hyperbola  <img alt="\mathrm{\frac{x^2}{b^2}-\frac{y^2}{a^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Bb%5E2%7D-%5Cfrac%7By%5E2%7D%7Ba%5

If hyperbola  \mathrm{\frac{x^2}{b^2}-\frac{y^2}{a^2}=1} passes through the foci of ellipse

\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}, then the square of eccentricity of hyperbola 

is

Option: 1

3


Option: 2

8


Option: 3

5


Option: 4

None of these


Answers (1)

best_answer

Let  \mathrm{e_1 \text { and } e_2}  be the eccentricities of ellipse and

hyperbola, respectively.

Since hyperbola passes through the foci of ellipse, we have 

                           \mathrm{b=a e_1}                                                       .......(1)

Also for ellipse  \mathrm{b^2=a^2\left(1-e_1^2\right)}                                       ........(2)

And for hyperbola  \mathrm{a^2=b^2\left(e_2^2-1\right)}                                 ........(3)

From (1) and (2) we get

\mathrm{2 e_1^2=1 \text { or } e_1=\frac{1}{\sqrt{2}}}

So, from (1) and (3) we get

\mathrm{2=e_2^2-1 \text { or } e_2=\sqrt{3} \Rightarrow e_2^2=3}

Posted by

Divya Prakash Singh

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