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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

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