Prove Theorem 5 in the case 0 −2 − 1).

THEOREM 5 Focus-Directrix Relationship Ellipse If 0 b>0 and c = a-b, then the ellipse () + () = satisfies Eq. (10) with F = (c,0),e=, and vertical directrix x = 2. = 1 Hyperbola If e > 1, then the set of points satisfying Eq. (10) is a hyperbola, and xy-coordinate axes can be chosen, and a, b defined, so that the hyperbola has eccentricity e and is in standard position with equation Conversely, if a, b>0 and c = a+ b, the hyperbola ()-() = satisfies Eq. (10) with F = (c, 0), e=, and vertical directrix x = 4.