Answer
Parabola: $e = 1$; Ellipse: $e = \sqrt{1 - \frac{b^2}{a^2}} < 1$ ($a > b$); Hyperbola: $e = \sqrt{1 + \frac{b^2}{a^2}} > 1$.
Explanation
Eccentricity measures the degree of deviation of a conic section from circularity ($e = 0$).
Home / Library / JEE Main Master Class: Calculus, Mechanics, Coordinate Geometry / Synthetic flashcard
Study flashcard from the deck JEE Main Master Class: Calculus, Mechanics, Coordinate Geometry.
Parabola: $e = 1$; Ellipse: $e = \sqrt{1 - \frac{b^2}{a^2}} < 1$ ($a > b$); Hyperbola: $e = \sqrt{1 + \frac{b^2}{a^2}} > 1$.
Eccentricity measures the degree of deviation of a conic section from circularity ($e = 0$).