Consider the equation. f(x)=2x3+10x−13 Therefore, f′(x)=6x2+10>0,∀x∈ℝ So, the function f(x) is strictly increasing. Thus, f(1)=2+10−13 =−1<0 And, f(2)=16+20−13 =23>0 So, the one and only one root e∈(1,2) where e is theeccentricity of the conic. As 1<e<2 This implies that the conic is a hyperbola.