Q.A parabola always has eccentricity $e=1$. Using the ellipse relation $b^2=a^2(1-e^2)$ and the hyperbola relation $b^2=a^2(e^2-1)$, explain why the ellipse's eccentricity must satisfy $0<e<1$ and the hyperbola's must satisfy $e>1$, and state what value of $e$ corresponds to a circle.

Every conic section -- circle, ellipse, parabola or hyperbola -- is obtained by intersecting a plane with a double right circular cone, and which curve results depends entirely on the angle the cutting plane makes with the cone's axis compared with the cone's own semi-vertical angle: a plane perpendicular to the axis gives a circle, one tilted between the axis-angle and the perpendicular gives an ellipse, one exactly parallel to a generator gives a parabola, and one steep enough to cut both nappes gives a two-branched hyperbola. If the same cutting plane is instead made to pass through the cone's vertex, each of these four ordinary conics collapses to a degenerate limiting case: the circle/ellipse family degenerates to a single point, the parabola family to a single straight line, and the hyperbola family to a pair of intersecting straight lines through the vertex. This geometric origin is what unifies four seemingly different curves into one family, and its eccentricity value …