If $5x + 9 = 0$ is the directrix of the hyperbola $16x^2 -9y^2 = 144,$ then its corresponding focus is

The straight line $x + y = \sqrt 2 p$ will touch the hyperbola $4{x^2} - 9{y^2} = 36$, if

Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellips $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then

$(A)$ the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$

$(B)$ a focus of the hyperbola is $(2,0)$

$(C)$ the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$

$(D)$ the equation of the hyperbola is $x^2-3 y^2=3$

The chord $ PQ $ of the rectangular hyperbola $xy = a^2$  meets the axis of $x$ at $A ; C $ is the mid point of $  PQ\ \& 'O' $ is the origin. Then the $ \Delta ACO$  is :

The minimum value of ${\left( {{x_1} - {x_2}} \right)^2} + {\left( {\sqrt {2 - x_1^2}  - \frac{9}{{{x_2}}}} \right)^2}$ where ${x_1} \in \left( {0,\sqrt 2 } \right)$ and ${x_2} \in {R^ + }$.

The locus of the point of intersection of any two perpendicular tangents to the hyperbola is a circle which is called the director circle of the hyperbola, then the eqn of this circle is