A line parallel to the straight line $2 x-y=0$ is tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point $\left(x_{1}, y_{1}\right) .$ Then $x_{1}^{2}+5 y_{1}^{2}$ is equal to 

The eccentricity of the hyperbola conjugate to ${x^2} - 3{y^2} = 2x + 8$ is

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

If the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ coincide with the foci of the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}},$ then $b^2$ is equal to

Let $H$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is

The minimum value of ${\left( {{x_1} - {x_2}} \right)^2} + {\left( {\sqrt {2 - x_1^2}  - \frac{9}{{{x_2}}}} \right)^2}$ where ${x_1} \in \left( {0,\sqrt 2 } \right)$ and ${x_2} \in {R^ + }$.