Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the midpoint of the line segment PQ is $\left( \frac{1}{2}, -\frac{1}{2} \right)$. Then the area of the triangle OPQ equals :

Let the ellipse $E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ and the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$ have the same foci. If $e$ and $L$

respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e+L)$ is :

Let PQ be a chord of the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$, perpendicular to the x -axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of the triangle OPQ is

Let the domain of the function $f(x)=\log _3 \log _5 \log _7\left(9 x-x^2-13\right)$ be the interval $(\mathrm{m}, \mathrm{n})$. Let the hyperbola $\frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ have eccentricity $\frac{\mathrm{n}}{3}$ and the length of the latus rectum $\frac{8 \mathrm{~m}}{3}$. Then $\mathrm{b}^2-\mathrm{a}^2$ is equal to :