$$P = \left( {1,0} \right)\,\,Q = \left( {h,k} \right)$$ Such that $${k^2} = 8h$$

Let $$\left( {\alpha ,\beta } \right)$$ be the midpoint of $$PQ$$

$$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$

$$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$

$${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$

$$ \Rightarrow {y^2} - 4x + 2 = 0.$$

as $$\angle FBF' = {90^ \circ }$$

$$ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}$$

$$\therefore$$ $${\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}$$

$$ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}$$

$$ \Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}$$



Also $${e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}$$

$$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$

$$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$

$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$

$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$

Equation of elhipe is

$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$

Solving equations of parabolas

$${y^2} = 4ax$$ and $${x^2} = 4ay$$

we get $$(0,0)$$ and $$(4a, 4a)$$

Substituting in the given equation of line

$$2bx+3cy+4d=0,$$

we get $$d=0$$

and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$