Radius = perpendicular distance from (1, -3) to the given line 3x - 4y - 5 = 0.
i.e. $\left|\frac{3+12-5}{\sqrt{5^2}}\right|=2.$

Radius = $\sqrt{(1–4{)}^2+(2-6{)}^2}=5$
Hence the area is given by $\pi r^2=25\pi$ sq. units.

Since the axis of parabola is y-axis 

$\therefore$ Equation of parabola $x^2=4ay$ 

Since it passes through (6 , -3)

$\therefore$ 36 = -12 a $\Rightarrow a=-3$ 

$y^2=4.\frac{1}{5}x;a=\frac{1}{5}$ . Focus is $(\frac{1}{5},0)$ and co-ordinates of latus rectum are $y^2=\frac{4}{25}\Rightarrow y=\pm \frac{2}{5}$ 

or end points of latus rectum are $(\frac{1}{5},\pm \frac{2}{5})$ . 

The given hyperbola is $x^2-y^2=9$
$\Rightarrow \frac{x^2}{9}-\frac{y^2}{9}=1$
$a=3,b=3,c=\sqrt{a^2+b^2}=3\sqrt{2}$
The equation of  the directrices are $x=\pm \frac{a^2}{c}$
Distance between directrices,
$d=\frac{2a^2}{c}=\frac{18}{3\sqrt{2}}=3\sqrt{2}$

The hyperbola is of the form 
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The transverse axis of the hyperbola is along the x-axis. Given that 2x + 3y-6=0 is a focal chord. Let (c,0) be the focus.
$\Rightarrow 2c+0-6=0\Rightarrow c=3Given,e=\frac{5}{4}a=\frac{c}{e}=\frac{12}{5}$

The length of the transverse axis is $2a=\frac{24}{5}$

The eccentricity of any conic is the ratio of its distances from the focus and directrix. Let d be the distance of the point P from the directrix. Since P lies on the ellipse,

$\frac{6}{d}=\frac{3}{5}\Rightarrow d=10$

For the given hyperbola, Length of the transverse axis = 2a
$\text{2a= 2cos t}\Rightarrow \text{a = cos t}$

Consider the ellipse

${\text{9x}}^2+{\text{16y}}^2=144\Rightarrow \frac{{\text{x}}^2}{16}+\frac{{\text{y}}^2}{9}=1\text{Focal length, c=}\sqrt{16-9}=\sqrt{7}$

This is the same as the focal length of the hyperbola.

Now, for the hyperbola,
${\text{b}}^2={\text{c}}^2-{\text{a}}^2{\text{b}}^2=7-{\text{cos}}^2\text{t}$

The equation of the hyperbola is
$\frac{{\text{x}}^2}{{\text{cos}}^2\text{t}}-\frac{{\text{y}}^2}{7-{\text{cos}}^2\text{t}}=1$

Consider the following hyperbola

$\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\frac{13x^2}{144}-\frac{13y^2}{25}=1a^2=\frac{144}{13};b^2=\frac{25}{13}\text{Focal length,}c=\sqrt{a^2+b^2}c=\sqrt{\frac{144+25}{13}}=\sqrt{\frac{169}{13}}=\sqrt{13}$

Also, this is the focus of the ellipse 
$\frac{x^2}{25}+\frac{y^2}{b^2}=1c^2=a^2-b^2\Rightarrow 13=25-b^2{b}^2=12$

Let the transverse axis be along the x-axis such that the hyperbola is of the form

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\text{Latus Rectum =}\frac{2b^2}{a}=4\sqrt{3}$
$\text{Also,}2a=2\sqrt{3}\Rightarrow a=\sqrt{3}$

$\frac{2b^2}{\sqrt{3}}=4\sqrt{3}\Rightarrow b^2=6$

The equation of the hyperbola is

$\frac{x^2}{3}-\frac{y^2}{6}=1$