Let $0 < \theta < \frac{\pi}{2}$. If the eccentricity of the hyperbola $\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1$ is greater than $2$,then the length of its latus rectum lies in the interval

On a rectangular hyperbola $x^2-y^2=a^2, a > 0$,three points $A, B, C$ are taken as follows: $A=(-a, 0)$; $B$ and $C$ are placed symmetrically with respect to the $X$-axis on the branch of the hyperbola not containing $A$. Suppose that the $\triangle ABC$ is equilateral. If the side length of the $\triangle ABC$ is $ka$,then $k$ lies in the interval

Let $P(\frac{\pi}{4}), Q(\frac{5 \pi}{4}), R(\frac{3 \pi}{4}), T(\frac{7 \pi}{4})$ be the points on the hyperbola $x^2-4y^2-4=0$ in the parametric form. Then the area of the quadrilateral $PQRT$ is (in square units) (in $\sqrt{2}$)

The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-

If the coordinates of a point are given by the equations $x = b \sec \phi$ and $y = a \tan \phi$,then its locus is

$A$ normal to the hyperbola $4x^2 - 9y^2 = 36$ meets the coordinate axes $x$ and $y$ at $A$ and $B$,respectively. If the parallelogram $OABP$ ($O$ being the origin) is formed,then the locus of $P$ is