An ellipse intersects the hyperbola $2{x^2} - 2{y^2} = 1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axis of the ellipse are along the coordinate axes, then
$\left( a \right)$ Equation of ellipse is ${x^2} + 2{y^2} = 2$
$\left( b \right)$ The foci of ellipse are $\left( { \pm 1,0} \right)$
$\left( c \right)$ Equation of ellipse is ${x^2} + 2{y^2} = 4$
$\left( d \right)$ The foci of ellipse are $\left( { \pm \sqrt 2 ,0} \right)$

Hint: In this particular question use the concept that if the equations ${a_1}{x^2} + {b_1}{y^2} = 1,{a_2}{x^2} + {b_2}{y^2} = 1$ intersects orthogonally then the condition is $\dfrac{1}{{{a_1}}} - \dfrac{1}{{{b_1}}} = \dfrac{1}{{{a_2}}} - \dfrac{1}{{{b_2}}}$ and use the concept that the relation between, eccentricity e, semi major axis a, and semi minor axis b in case of hyperbola is given as ${b^2} = {a^2}\left( {{e^2} - 1} \right)$ so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
Given equation of hyperbola
$2{x^2} - 2{y^2} = 1$
Let the equation of ellipse is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]
Now as we know that that if the equations ${a_1}{x^2} + {b_1}{y^2} = 1,{a_2}{x^2} + {b_2}{y^2} = 1$ intersects orthogonally then the condition is $\dfrac{1}{{{a_1}}} - \dfrac{1}{{{b_1}}} = \dfrac{1}{{{a_2}}} - \dfrac{1}{{{b_2}}}$................ (1)
So on comparing with equation of ellipse,
$ \Rightarrow {a_1} = \dfrac{1}{{{a^2}}},{b_1} = \dfrac{1}{{{b^2}}}$
And on comparing with equation of hyperbola we have,
$ \Rightarrow {a_2} = 2,{b_2} = - 2$
Now from equation (1) we have,
$ \Rightarrow {a^2} - {b^2} = \dfrac{1}{2} - \left( { - \dfrac{1}{2}} \right) = \dfrac{1}{2} + \dfrac{1}{2}$
$ \Rightarrow {a^2} - {b^2} = 1$................... (2)
Now as we know that the relation between, eccentricity ${e_3}$, semi major axis ${a_3}$, and semi minor axis ${b_3}$ in case of hyperbola is given as $b_3^2 = a_3^2\left( {e_3^2 - 1} \right)$.
Now the standard equation of hyperbola is \[\dfrac{{{x^2}}}{{a_3^2}} - \dfrac{{{y^2}}}{{b_3^2}} = 1\] now compare it with given equation of hyperbola $2{x^2} - 2{y^2} = 1$ we have,
$ \Rightarrow a_3^2 = \dfrac{1}{2},b_3^2 = \dfrac{1}{2}$
Therefore, $b_3^2 = a_3^2\left( {e_3^2 - 1} \right)$ become,
$ \Rightarrow \dfrac{1}{2} = \dfrac{1}{2}\left( {e_3^2 - 1} \right)$
$ \Rightarrow 1 = \left( {e_3^2 - 1} \right)$
$ \Rightarrow e_3^2 = 2$
$ \Rightarrow {e_3} = \sqrt 2 $
Now it is given that the eccentricity of the ellipse is reciprocal of that of the hyperbola.
So ${e_1} = \dfrac{1}{{{e_3}}} = \dfrac{1}{{\sqrt 2 }}$, where ${e_1}$ is the eccentricity of the ellipse.
Now as we know that the relation between, eccentricity e, semi major axis a, and semi minor axis in case of ellipse is given as ${b^2} = {a^2}\left( {1 - e_1^2} \right)$.
$ \Rightarrow {b^2} = {a^2}\left( {1 - {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \right) = {a^2}\left( {1 - \dfrac{1}{2}} \right) = \dfrac{{{a^2}}}{2}$
Now substitute this value in equation (2) we have,
$ \Rightarrow {a^2} - \dfrac{{{a^2}}}{2} = 1$
$ \Rightarrow \dfrac{{{a^2}}}{2} = 1$
$ \Rightarrow {a^2} = 2$
$ \Rightarrow a = \sqrt 2 $
Now from equation (2), we have
$ \Rightarrow {a^2} - {b^2} = 1$
$ \Rightarrow 2 - {b^2} = 1$
$ \Rightarrow {b^2} = 1$
So the equation of ellipse becomes
\[ \Rightarrow \dfrac{{{x^2}}}{2} + \dfrac{{{y^2}}}{1} = 1\]
\[ \Rightarrow {x^2} + 2{y^2} = 2\]
Mow as we know that the foci of an ellipse is given as $\left( { \pm a{e_1},0} \right)$
Therefore, foci of an ellipse = $\left( { \pm \sqrt 2 \times \dfrac{1}{{\sqrt 2 }},0} \right) = \left( { \pm 1,0} \right)$
So this is the required answer.
Hence options (a) and (b) are the correct answers.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the condition of orthogonality which is stated above, and always recall the standard equation of ellipse as well as a hyperbola, and always recall the relation between the eccentricity e, semi-major axis a, and semi-minor axis b in case of hyperbola as well as in case of an ellipse.