$$\begin{aligned} &\text { Given that, foci of the ellipse are }(0, \pm b e)\\ &\begin{array}{ll} \because & b e \equiv 1 \\ \therefore & \text { Length of minor axis, } 2 a=1 \Rightarrow a=1 / 2 \end{array} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} e^2 & =1-\frac{a^2}{b^2} \\ \Rightarrow \quad(b e)^2 & =b^2-a^2 \Rightarrow 1=b^2-\frac{1}{4} \\ \Rightarrow \quad 1+\frac{1}{4} & =b^2 \Rightarrow \frac{5}{4}=b^2 \end{aligned}\\ &\text { So, the equation of ellipse is }\\ &\frac{x^2}{\frac{1}{4}}+\frac{y^2}{5 / 4}=1 \Rightarrow \frac{4 x^2}{1}+\frac{4 y^2}{5}=1 \end{aligned}$$

Given that, focus at $F(-1,-2)$ and directrix is $x-2 y+3=0$

Let any point on the parabola be $(x, y)$.

$$\begin{array}{ll} \therefore & \quad P F=\left|\frac{x-2 y+3}{\sqrt{1+4}}\right| \\ \Rightarrow & (x+1)^2+(y+2)^2=\frac{(x-2 y+3)^2}{5} \\ \Rightarrow & 5\left[x^2+2 x+1+y^2+4 y+4\right]=x^2+4 y^2+9-4 x y-12 y+6 x \\ \Rightarrow & 4 x^2+y^2+4 x+32 y+16=0 \end{array}$$

Let the equation of the hyperbola be $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Then vertices $=(0, \pm b)=(0, \pm 6)$

$$\begin{array}{llrl} \therefore & b =6 \text { and } e=5 / 3 \\ \because & e =\sqrt{1+\frac{a^2}{b^2}} \Rightarrow \frac{25}{9}=1+\frac{a^2}{36} \\ \Rightarrow & \frac{25-9}{9} =\frac{a^2}{36} \Rightarrow 16=\frac{a^2}{4} \Rightarrow a^2=48 \end{array}$$

So, the equation of hyperbola is,

$$\begin{aligned} \frac{-x^2}{48}+\frac{y^2}{36} & =1 \Rightarrow \frac{y^2}{36}-\frac{x^2}{48}=1 \\ \text { Foci } & =(0, \pm \text { be })=\left(=0, \pm \frac{5}{3} \times 6\right)=(0, \pm 10) \end{aligned}$$