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माना दीर्घवृत्त $\frac{x^2}{16}+\frac{y^2}{7}=1$ तथा अतिपरवलय $\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}$ की नाभियाँ सम्पाती हैं। तो अतिपरवलय के नाभिलंब जीवा की लंबाई है :
$\frac{32}{9}$
$\frac{18}{5}$
$\frac{27}{4}$
$\frac{27}{10}$
Solution
Ellipse : $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$
Eccentricity $=\sqrt{1-\frac{7}{16}}=\frac{3}{4}$
Foci $\equiv(\pm a \quad e, 0) \equiv(\pm 3,0)$
Hyperbola : $\frac{x^{2}}{\left(\frac{144}{25}\right)}-\frac{y^{2}}{\left(\frac{\alpha}{25}\right)}=1$
Eccentricity $=\sqrt{1+\frac{\alpha}{144}}=\frac{1}{12} \sqrt{144+\alpha}$
Foci $\equiv(\pm a e, 0) \equiv\left(\pm \frac{12}{5} \cdot \frac{1}{12} \sqrt{144+\alpha}, 0\right)$
If foci coincide then $3=\frac{1}{5} \sqrt{144+\alpha} \Rightarrow \alpha=81$
Hence, hyperbola is $\frac{x^{2}}{\left(\frac{12}{5}\right)^{2}}-\frac{y^{2}}{\left(\frac{9}{5}\right)^{2}}=1$
Length of latus rectum $=2 \cdot \frac{81 / 25}{12 / 5}=\frac{27}{10}$
