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**Hint:**We can use formula for eccentricity and focus and hyperbola because in the given question we have a relation between the eccentricity of ellipse and hyperbola. Then we can use the focus of the ellipse to get the final answer.

**Complete step-by-step solution:**In given question equation of ellipse is ${{x}^{2}}+4{{y}^{2}}=4$.

On dividing both sides from 4 to write it in standard form.

$\Rightarrow \dfrac{{{x}^{2}}+4{{y}^{2}}}{4}=\dfrac{4}{4}$

$\Rightarrow \dfrac{{{x}^{2}}}{4}+\dfrac{4{{y}^{2}}}{4}=1$

$\Rightarrow \dfrac{{{x}^{2}}}{4}+{{y}^{2}}=1$

On comparing with $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

$\Rightarrow {{a}^{2}}=4,\,{{b}^{2}}=1$ where a is half of the major axis and b is half of the minor axis.

To calculate eccentricity of ellipse we can use ${{b}^{2}}={{a}^{2}}(1-{{e}^{2}})$

We can arrange it as

$\Rightarrow {{b}^{2}}={{a}^{2}}(1-{{e}^{2}})$

$\Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}=1-{{e}^{2}}$

$\Rightarrow {{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$

On substitute ${{a}^{2}}=4,\,{{b}^{2}}=1$

$\Rightarrow {{e}^{2}}=1-\dfrac{1}{4}$

$\Rightarrow {{e}^{2}}=\dfrac{3}{4}$

$\Rightarrow e=\pm \sqrt{\dfrac{3}{4}}$

$\Rightarrow e=\dfrac{\sqrt{3}}{2}\,or\,\dfrac{-\sqrt{3}}{2}$

But eccentricity of the ellipse always lies between 0 to 1.

Hence

$\Rightarrow e=\dfrac{\sqrt{3}}{2}$

As given in question eccentricity of a hyperbola is reciprocal of the eccentricity of the ellipse.

Let the equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}_{1}}^{2}}-\dfrac{{{y}^{2}}}{{{b}_{1}}^{2}}=1$

Hence eccentricity of hyperbola is $e=\dfrac{2}{\sqrt{3}}$

To calculate eccentricity of hyperbola we can use ${{b}_{1}}^{2}={{a}_{1}}^{2}({{e}^{2}}-1)$ where ${{a}_{1}}$ is the transverse axis of hyperbola and ${{b}_{1}}$ is conjungate axis of hyperbola

On substituting $e=\dfrac{2}{\sqrt{3}}$

$\Rightarrow {{b}_{1}}^{2}={{a}_{1}}^{2}\left( {{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-1 \right)$

$\Rightarrow {{b}_{1}}^{2}={{a}_{1}}^{2}\left( \dfrac{4}{3}-1 \right)$

$\Rightarrow {{b}_{1}}^{2}={{a}_{1}}^{2}\left( \dfrac{4-3}{3} \right)$

$\Rightarrow {{b}_{1}}^{2}=\dfrac{{{a}_{1}}^{2}}{3}$………………………………………….(i)

Co-ordinate of focus of ellipse is $\left( \pm ae,0 \right)$ if ${{a}^{2}}>{{b}^{2}}$.

From equation of ellipse $a=2,\,e=\dfrac{\sqrt{3}}{2}$

$\Rightarrow ae=2\times \dfrac{\sqrt{3}}{2}$

$\Rightarrow ae=\sqrt{3}$

Hence co-ordinate of focus of ellipse is $\left( \pm \sqrt{3},0 \right)$

As given, equation of hyperbola passes through focus of ellipse. Hence it will satisfy equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}_{1}}^{2}}-\dfrac{{{y}^{2}}}{{{b}_{1}}^{2}}=1$

$\Rightarrow \dfrac{{{\left( \sqrt{3} \right)}^{2}}}{{{a}_{1}}^{2}}-\dfrac{{{\left( 0 \right)}^{2}}}{{{b}_{1}}^{2}}=1$

$\Rightarrow \dfrac{3}{{{a}_{1}}^{2}}-0=1$

$\Rightarrow \dfrac{3}{{{a}_{1}}^{2}}=1$

$\Rightarrow {{a}_{1}}^{2}=3$

On substituting ${{a}_{1}}^{2}=3$ in equation (i) ${{b}_{1}}^{2}=\dfrac{{{a}_{1}}^{2}}{3}$

$\Rightarrow {{b}_{1}}^{2}=\dfrac{3}{3}$

$\Rightarrow {{b}_{1}}^{2}=1$

Hence equation of hyperbola is by substituting value of ${{a}_{1}}^{2}$ and ${{b}_{1}}^{2}$

$\Rightarrow \dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{1}=1$

We can simplify by taking L.C.M

$\Rightarrow \dfrac{{{x}^{2}}-3{{y}^{2}}}{3}=1$

$\Rightarrow {{x}^{2}}-3{{y}^{2}}=3$

In general focus of hyperbola is $\left( \pm {{a}_{1}}e,0 \right)$ if ${{a}_{1}}^{2}>{{b}_{1}}^{2}$.

For hyperbola $e=\dfrac{2}{\sqrt{3}},{{a}_{1}}=\sqrt{3}$

$\Rightarrow {{a}_{1}}e=\sqrt{3}\times \dfrac{2}{\sqrt{3}}$

$\Rightarrow {{a}_{1}}e=2$

Hence focus of hyperbola is $\left( \pm 2,0 \right)$.

**So option b and d is correct.**

**Note:**In general if equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ then eccentricity of ellipse can be calculated from relation ${{b}^{2}}={{a}^{2}}(1-{{e}^{2}})$. Co-ordinate of focus of ellipse is $\left( \pm ae,0 \right)$ if ${{a}^{2}}>{{b}^{2}}$.

If equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}_{1}}^{2}}-\dfrac{{{y}^{2}}}{{{b}_{1}}^{2}}=1$ then eccentricity of hyperbola can be calculated from relation ${{b}_{1}}^{2}={{a}_{1}}^{2}({{e}^{2}}-1)$.

In general eccentricity(e) of the conic section defines its shape and it is a non negative real number.

For ellipse,$0 < e < 1$

For hyperbola, $e >1$

In equations of ellipse and hyperbola if any variable is common then it has to be represented separately to avoid any error arising due to common variable used in the equation

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