Given that, eccentricity $=\frac{5}{8}$, i.e., $e=\frac{5}{8}$

Let equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,

Since the foci of this ellipse is ( $\pm \mathrm{ae}, 0$ ).

$\therefore \quad$ Distance between foci $=\sqrt{(a e+a e)^2}$

$$\begin{array}{lll} \Rightarrow & 2 \sqrt{a^2 e^2}=10 & {[\because \text { distance between its foci }=10]} \\ \Rightarrow & \sqrt{a^2 e^2}=5 \\ \Rightarrow & a^2 e^2=25 \\ \Rightarrow & a^2=\frac{25 \times 64}{25} \\ \therefore & a=8 \end{array}$$

We know that,

$$\begin{array}{ll} \Rightarrow & b^2=a^2\left(1-e^2\right) \\ \Rightarrow & b^2=64\left(1-\frac{25}{64}\right) \\ \Rightarrow & b^2=64\left(\frac{64-25}{64}\right) \\ & b^2=39 \end{array}$$

$\therefore \quad$ Length of latusrectum of ellipse $=\frac{2 b^2}{a}=2\left(\frac{39}{8}\right)=\frac{39}{4}$

Given that, $e=2 / 3$ and latusrectum $=5$ i.e.,

$$\begin{aligned} \frac{2 b^2}{a} & =5 \Rightarrow b^2=\frac{5 a}{2} \\ \text{We know that,}\quad b^2 & =a^2\left(1-e^2\right) \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \frac{5 a}{2}=a^2\left(1-\frac{4}{9}\right) \\ \Rightarrow & \frac{5}{2}=\frac{5 a}{9} \Rightarrow a=9 / 2 \Rightarrow a^2=\frac{81}{4} \\ \Rightarrow & b^2=\frac{5 \times 9}{2 \times 2}=\frac{45}{4} \end{array}$$

So, the required equation of the ellipse is $\frac{4 x^2}{81}+\frac{4 y^2}{45}=1$.

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 $y^2=4 a x$

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