If e and $e'$ be the eccentricities of the two conics S=0 and $S' = 0$ and if ${e^2} + {e'^2} = 3$ then both $S{\text{ and S' }}$ can be
(a) hyperbolas
(b) ellipses
(c) parabolas
(d) none of these

Hint– In this question use the basic definition of eccentricity for parabola and ellipse. Then recall that the eccentricity in case of parabola is equal to one, similarly in case of ellipse it is lesser than 1 and in case of hyperbola it is greater than 1. Now consider this to satisfy random values of eccentricities to the given equation of eccentricity that is ${e^2} + {\left( {e'} \right)^2} = 3$.

Complete step-by-step answer:
Given data:
Equation of conic sections are
$S = 0{\text{ and }}S' = 0$
Eccentricities of the conic sections are $e{\text{ and }}e'$ respectively.
And the given equation is
${e^2} + {\left( {e'} \right)^2} = 3$
Now the eccentricities of the conic sections are given below:
$\left( i \right)$ Parabola: In a parabola the ratio of the distance (${S_{FP}}$) from the focus of the parabola to any general point (x, y) on the parabola to the perpendicular distance (${S_{PD}}$) from this general point on the directrix of the parabola is called as eccentricity (e) of the parabola and this is equal to 1.
$ \Rightarrow \dfrac{{{S_{FP}}}}{{{S_{PD}}}} = e = 1$
$\left( {ii} \right)$ Ellipse: Definition is same as above only difference is that the eccentricity is always lesser than 1.
$ \Rightarrow \dfrac{{{S_{FP}}}}{{{S_{PD}}}} = e < 1$
$\left( {iii} \right)$ Hyperbola: Definition is the same as above, the only difference is that the eccentricity is always greater than 1.
$ \Rightarrow \dfrac{{{S_{FP}}}}{{{S_{PD}}}} = e > 1$
Now in a parabola eccentricity is always 1 so it never satisfies the given equation as if e = 1 and e’ = 1
$ \Rightarrow {e^2} + {\left( {e'} \right)^2} = 1 + 1 = 2$
Which does not satisfy the given equation.
Now in ellipse eccentricity is always less than 1 so it also never satisfies the given equation. As less than 1 whole square is less than the previous value and the sum of these values can never be equal to three.
Now in hyperbola the eccentricity is always greater than 1 so it can be satisfied by the given equation.
Let the eccentricity of the first hyperbola be $e = \sqrt {1.2} $ and the eccentricity of the second hyperbola be $e' = \sqrt {1.8} $.
Therefore, ${e^2} + {\left( {e'} \right)^2} = {\left( {\sqrt {1.2} } \right)^2} + {\left( {\sqrt {1.8} } \right)^2} = 1.2 + 1.8 = 3$
Which satisfies the given equation.
Hence both the conic sections can be hyperbolas.
So this is the required answer.
Hence option (A) is the correct answer.

Note – Hit and trial of the values of eccentricity is the only method to solve problems of this kind as it has a single equation involving two variables that is two different eccentricities. One important concept about eccentricity is that it is always non-negative that eccentricity can’t be negative and eccentricity uniquely characterizes the shape of a conic section.