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  • Let the foci of the ellipse <img alt="\mathrm{\frac{x^{2}}{16}+\frac{y^{2}}{7}=1}" src="/latex-image/?%5Cmathrm%7B%5Cfrac%7Bx%5E%7B2%7D%7D%7B16%7D&plus;%5Cfrac%7By%5E%7B2%7D%7D%7B7%7D%3D1%7D" /

Let the foci of the ellipse \mathrm{\frac{x^{2}}{16}+\frac{y^{2}}{7}=1} and the hyperbola \mathrm{\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}} coincide. Then the length of the latus rectum of the hyperbola is:

Option: 1

\frac{32}{9}


Option: 2

\frac{18}{5}


Option: 3

\frac{27}{4}


Option: 4

\frac{27}{10}


Answers (1)

best_answer

For the eclipse are

\mathrm{=4 \sqrt{1-\frac{7}{16}}} \\

\mathrm{=4 \times \frac{3}{4}} \\

=3

For hyperbola ,ae should be 3

\mathrm{\frac{x^{2}}{\left(\frac{144}{25}\right)}-\frac{y^{2}}{\left(\frac{\alpha}{25}\right)}=1} \\

\mathrm{\Rightarrow \quad \frac{12}{5} \cdot \sqrt{1+\frac{\alpha}{144}}=3} \\

\mathrm{\Rightarrow \quad \sqrt{\frac{144+\alpha}{144}}=\frac{15}{12} }\\

\mathrm{\Rightarrow \quad144+\alpha =225}

\mathrm{\Rightarrow \quad \alpha=81}

Lotus rectum    \mathrm{=\frac{2 b^{2}}{a}} \\

                      \mathrm{=\frac{2 \cdot 81}{25 \cdot \frac{12}{5}}=\frac{27}{10}}

Hence correct option is 4  

Posted by

mansi

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