If the product of the perpendicular distances from any point on the hyperbola $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\,\, = \,1$ of eccentricity $e =\sqrt 3 \,$ from its asymptotes is equal to $6$, then the length of the transverse axis of the hyperbola is

The equation of the hyperbola whose directrix is $x + 2y = 1$, focus $(2, 1)$ and eccentricity $2$ will be

For hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1$ which of the following remain constant if $\alpha$ varies

The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$

$2.$ Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

$(A)$ $2 x-\sqrt{5} y-20=0$ $(B)$ $2 x-\sqrt{5} y+4=0$

$(C)$ $3 x-4 y+8=0$ $(D)$ $4 x-3 y+4=0$

$2.$ Equation of the circle with $\mathrm{AB}$ as its diameter is

$(A)$ $x^2+y^2-12 x+24=0$ $(B)$ $x^2+y^2+12 x+24=0$

$(C)$ $\mathrm{x}^2+\mathrm{y}^2+24 \mathrm{x}-12=0$ $(D)$ $x^2+y^2-24 x-12=0$

Give hte answer question $1, 2$

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is

Let the tangent to the parabola $y^2=12 x$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$.Then the square of distance of the point $(6,-4)$from the normal to the hyperbola $\alpha^2 x^2-9 y^2=9 \alpha^2$at its point $(\alpha-1, \alpha+2)$ is equal to $........$.