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  • Let the hyperbola <img alt="\mathrm{H: \frac{x^{2}}{a^{2}}-y^{2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BH%3A%20%5Cfrac%7Bx%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D-y%5E%7B2%7D%3D

Let the hyperbola \mathrm{H: \frac{x^{2}}{a^{2}}-y^{2}=1} and the ellipse \mathrm{E: 3 x^{2}+4 y^{2}=12} be such that the length of latus rectum of \mathrm{H} is equal to the length of latus rectum of \mathrm{E}. If  and \mathrm{e_{H}}are the eccentricities of \mathrm{H} and \mathrm{E} respectively, then the value of \mathrm{12\left(e_{H}^{2}+e_{E}^{2}\right)} is equal to______

Option: 1

42


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} &H: \frac{x^{2}}{a^{2}}-y^{2}=1\\ &\text { E: } \quad 3 x^{2}+4 y^{2}=12\\ \end{aligned}

\begin{aligned} &(L R)_{H}=(L R)_{E} \\&2 \times \frac{1}{a}=2 \times \frac{3}{2}\\ &a=\frac{2}{3}\\\end{aligned}               \mathrm{\begin{aligned} & \stackrel{\text { AlSO }}{1=a^{2}\left(e_{H}^{2}-1\right)}\\ &e_{{H}}{ }^{2}=\frac{13}{4}\\ &\text { \& }\\ &3=4\left(1-e_{E}{ }^{2}\right)\\ &e_{E^{2}}=\frac{1}{4}\\ \end{aligned}}

\mathrm{\begin{aligned} & 12 (e_{H}^{2}+e_{E}^{2})\\ &=12\left (\frac{14}{4} \right )=42 \end{aligned}}

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