The equation of ellipse is $\frac{x^2}{36}+\frac{y^2}{20}=1$.

On comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get

$$\begin{array}{r} & \text { We know that, } & a =6, b=2 \sqrt{5} \\ \Rightarrow & b^2 =a^2\left(1-e^2\right) \\ \Rightarrow & 20 =36\left(1-e^2\right) \\ \therefore & \frac{20}{36} =1-e^2 \\ & e =\sqrt{1-\frac{20}{36}}=\sqrt{\frac{16}{36}} \\ & E =\frac{4}{6}=\frac{2}{3} \end{array}$$

$$\begin{aligned} & \text { Now, } \quad \text { directrices }=\left(+\frac{a}{e},-a / e\right) \\ & \therefore \quad \frac{a}{e}=\frac{\frac{6}{2}}{3}=\frac{6 \times 3}{2}=9 \\ & \text { and } \quad -\frac{a}{e}=-9 \end{aligned}$$

$\therefore$ Distance between the directrices $=|9-(-9)|=18$

Given parabola is $y^2=8 x\quad \text{... (i)}$

On comparing this parabola to the $y^2=4 a x$, we get

$$8 x=4 a x \Rightarrow a=2$$

$\therefore$ Focal distance $=|x+a|=4$

$$\begin{aligned} & \Rightarrow \quad|x+2|=4 \\ & \Rightarrow \quad x+2= \pm 4 \\ & \Rightarrow \quad x=2,-6 \\ & \text { But } \\ & x \neq-6 \\ & \text { For } x=2 \text {, } \\ & y^2=8 \times 2 \\ & \therefore \quad y^2=16 \Rightarrow y= \pm 4 \end{aligned}$$

So, the points are $(2,4)$ and $(2,-4)$.

Given equation of the parabola is $y^2=4 a x$

$$\begin{aligned} &\text { Let the coordinates of any point } \left.P \text { on the parabola be ( } a t^2, 2 a t\right) \text {. }\\ &\begin{aligned} & \text { In } \triangle P O A, \quad \tan \theta=\frac{2 a t}{a t^2}=\frac{2}{t} \\ & \Rightarrow \quad \tan \theta=\frac{2}{t} \Rightarrow t=2 \cot \theta \\ & \therefore \quad \text { length of } O P=\sqrt{\left(0-a t^2\right)^2+(0-2 a t)^2} \\ & =\sqrt{a^2 t^4+4 a^2 t^2} \\ & =a t \sqrt{t^2+4} \\ & =2 a \cot \theta \sqrt{4 \cot ^2 \theta+4} \\ & =4 a \cot \theta \sqrt{1+\cot ^2 \theta} \\ & =4 a \cot \theta \cdot \operatorname{cosec} \theta \\ & =\frac{4 a \cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta}=\frac{4 a \cos \theta}{\sin ^2 \theta} \end{aligned} \end{aligned}$$

Given that the coordinates, vertex of the parabola $(0,4)$ and focus of the parabola $(0,2)$.

$$ \begin{aligned} &\text { By definition of the parabola, } \quad P B=P F\\ &\begin{array}{cc} \Rightarrow & \left|\frac{0+y-6}{\sqrt{0+1}}\right|=\sqrt{(x-0)^2+(y-2)^2} \\ \Rightarrow & |y-6|=\sqrt{x^2+y^2-4 y+4} \\ \Rightarrow & x^2+y^2-4 y+4=y^2+36-12 y \\ \Rightarrow & x^2+8 y=32 \end{array} \end{aligned}$$

Given that, line $y=m x+1$ is tangent to the parabola $y^2=4 x$.

$\therefore\quad y =m x+1\quad\text{... (i)}$

$\text{and}\quad y^2 =4 x\quad \text{... (ii)}$

From Eqs. (i) and (ii),

$$\begin{array}{r} & m^2 x^2+2 m x+1 =4 x \\ \Rightarrow & m^2 x^2+2 m x-4 x+1 =0 \\ \Rightarrow & m^2 x^2+x(2 m-4)+1 =0 \\ \Rightarrow & (2 m-4)^2-4 m^2 \times 1 =0 \\ \Rightarrow & 4 m^2+16-16 m-4 m^2 =0 \\ \Rightarrow & 16 m =16 \\ \therefore & m=1 \end{array}$$