Maximum length of chord of the ellipse $\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1$, such that eccentric angles of its extremities differ by $\frac{\pi }{2}$ is 

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1 \,\,and\,\,\left| {b - 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \le 36} \right\}$ then : . . . . .

Let $P$ is any point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ . $S_1$ and $S_2$ its foci then maximum area of $\Delta PS_1S_2$ is (in square units)

Let $P(a\sec \theta ,\;b\tan \theta )$ and $Q(a\sec \varphi ,\;b\tan \varphi )$, where $\theta + \phi = \frac{\pi }{2}$, be two points on the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$, then $k$ is equal to

If the area of the auxiliary circle of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a > b} \right)$ is twice the area of the ellipse, then the eccentricity of the ellipse is

Let the length of the latus rectum of an ellipse with its major axis long $x -$ axis and center at the origin, be $8$. If the distance between the foci of this ellipse is equal to the length of the length of its minor axis, then which one of the following points lies on it?