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

$${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$

$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$

$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$

$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$

$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola

$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$

$$\therefore$$ $$e = {3 \over 4}$$

Then $${b^2} = {a^2}\left( {1 - {e^2}} \right)$$

$$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$