Let the foci of the ellipse frac{x^{2}}{16}+f | vedclass

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Question

Ellipse : $\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$

Eccentricity $=\sqrt{1-\frac{7}{16}}=\frac{3}{4}$

Foci $\equiv(\pm a \quad e, 0) \equiv(\pm 3,0)$

Vedclass · Accepted Answer

$\frac{27}{10}$

Vedclass · Answer

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

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

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