Prove Theorem 5 in the case 0 2 1). THEOREM 5 Focus-Directrix Relationship Ellipse If 0

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

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THEOREM 5 Focus-Directrix Relationship Ellipse • If 0 b>0 and c = √a²-b², then the ellipse ()² + ()² = 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 = √a²+ b², the hyperbola ()²-()² = ¹ satisfies Eq. (10) with F = (c, 0), e=, and vertical directrix x = 4.