$\text{ If the eccentricity of the hyperbola }{x}^2-y^2{\sec }^2\theta =5\text{ is }\sqrt{3}\text{ times the eccentricity of the }$

$\text{ ellipse }{x}^2{\sec }^2\theta +y^2=25,\text{ and smallest positive value of }\theta \text{ is }\frac{\pi }{P}\text{ then twice the value of 'p'is}$

$\text{ Eccentricity of the hyperbola }{x}^2-y^2{\sec }^2\theta =5\text{ is }$

equation is $\frac{x^2}{5}-\frac{y^2}{{\displaystyle \frac{5}{se{c}^2\theta }}}=1$

$\begin{array}{l}{e}_1=\sqrt{\frac{1+{\sec }^2\theta }{{\sec }^2\theta }}=\sqrt{1+{\cos }^2\theta }\\ \text{ Eccentricity of the ellipse }{x}^2{\sec }^2\theta +y^2=25\text{ is }\\ \frac{x^2}{{\displaystyle \frac{25}{se{c}^2\theta }}}+\frac{y^2}{25}=1\\ e_2=\sqrt{\frac{{\sec }^2\theta -1}{{\sec }^2\theta }}=|\sin \theta |\text{ Given }{e}_1=\sqrt{3}{e}_2\\ \Rightarrow 1+{\cos }^2\theta =3{\sin }^2\theta \Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}\\ \therefore \text{ Least positive value of }\theta \text{ is }\frac{\pi }{4}\therefore P=4\Rightarrow 2P=8\end{array}$