For the hyperbola $H : x^{2} - y^{2} = 1$ and the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $a > b > 0$,let $(1)$ the eccentricity of $E$ be the reciprocal of the eccentricity of $H$,and $(2)$ the line $y = \sqrt{\frac{5}{2}} x + K$ be a common tangent of $E$ and $H$. Then $4(a^{2} + b^{2})$ is equal to:

For $0 < \theta < \pi / 2$,if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$,then the value of $\theta$ is:

If the curves $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ cut each other orthogonally, then $k=$

Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$.  Let the sets  $S_1 = \{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas} \} = (\alpha, \beta),$ and $S_2 = \{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively} \} = (\gamma, \infty).$ Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4\sqrt{6}}{3}$. Let $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of $E$, and whose length of latus rectum is equal to the length of minor axis of $E$. Then the distance between the foci of $H$ is :

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: