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

The equation of the tangent to the hyperbola $2{x^2} – 3{y^2} = 6$ which is parallel to the line $y = 3x + 4$, is

Eccentricity of the rectangular hyperbola $\int_0^1 {{e^x}\left( {\frac{1}{x} – \frac{1}{{{x^3}}}} \right)} \;dx$ is

On a rectangular hy perbola $x^2-y^2= a ^2, a >0$, 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 $\triangle A B C$ is equilateral. If the side length of the $\triangle A B C$ is $k a$, then $k$ lies in the interval

A normal to the hyperbola, $4x^2 – 9y^2\, = 36$ meets the co-ordinate 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