The eccentricity of the conjugate hyperbola of the hyperbola ${x^2} - 3{y^2} = 1$, is

Length of latus rectum of hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 4\,is\,\left( {\alpha  \ne \frac{{n\pi }}{2},n \in I} \right)$

The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is

Let $H : \frac{ x ^{2}}{ a ^{2}}-\frac{y^{2}}{ b ^{2}}=1$, a $>0, b >0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2 \sqrt{2}+\sqrt{14})$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then value of $a^{2}+b^{2}$ is equal to

If the tangent on the point $(2\sec \phi ,\;3\tan \phi )$ of the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{9} = 1$ is parallel to $3x - y + 4 = 0$, then the value of $\phi$ is ............ $^o$

Let a line $L_{1}$ be tangent to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ and let $L_{2}$ be the line passing through the origin and perpendicular to $L _{1}$. If the locus of the point of intersection of $L_{1}$ and $L_{2}$ is $\left(x^{2}+y^{2}\right)^{2}=$ $\alpha x^{2}+\beta y^{2}$, then $\alpha+\beta$ is equal to