Question

Let the foci of thhe 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:

Answer 1

Ellipse: $\frac{x^2}{16}+\frac{y^2}{7}=1$
Eccentricity $=\sqrt{1-\frac{7}{16}}=\frac{3}{4}$
Foci $\equiv (\pm ae,0)\equiv (\pm 3,0)$
Hyperbola: $\frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{\alpha }{25}\right)}=1$
Eccentricity $=\sqrt{1+\frac{\alpha }{144}}=\frac{1}{12}\sqrt{144+\alpha }$
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 $=2.\frac{81/25}{12/5}=\frac{27}{10}$