Correct Answer - Option 3 : hyperbolas
Concept:
For an Ellipse the eccentricity is less than 1
For a Hyperbola the eccentricity is greater than 1
Calculation:
\(e_1^2 + e_2^2 = 3\)
e1 and e2 are eccentricities of same type of conic.
Let e1 = e2
⇒ 2e12 = 3
⇒ e12 = 1.5
⇒ e1 = √1.5 > 1
∴ Both conics S1 and S2 is Hyperbolas