Answer
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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
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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