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Prove some conditions by which a conic section is an ellipse or a parabola

Consider a conic section C with eccentricity e, a focus at the origin, and with a vertical directrix L which is a distance d to the left of F.

  1. Prove that if C is an ellipse or parabola then every point of C lies to the right of L and satisfies the equation (in polar coordinates)

        \[ r = \frac{ed}{1 - e \cos \theta}. \]

  2. Prove that if the conic section C is a hyperbola, point on the right branch satisfy the equation in part (a) while points on the left branch satisfy

        \[ r = \frac{-ed}{1 + e \cos \theta}. \]


  1. Proof. If C is a conic section with eccentricity e and focus at the origin, then we know the points on C are given by those points X such that

        \[ \lVert X \rVert = e | X \cdot N + d| \]

    where L is the directrix at a distance d to the left of F. So, if X has polar coordinates r and \theta, we have \lVert X \rVert = r and X \cdot N = r \cos \theta. Therefore,

        \[ r = e | r \cos \theta + d | \quad \implies \quad r = \frac{ed}{1 - e \cos \theta}. \]

    if X lies to the right of the directrix L (so | r \cos \theta + d | = r \cos \theta + d since r \cos \theta < d in this case), and since r > 0, we must have e \leq 1. \qquad \blacksquare

  2. Proof. If C is a hyperbola, then we know e > 1 and points on the left branch still satisfy

        \[ r = \frac{ed}{1 - e \cos \theta} \]

    just as in part (a). For the points on C to the right of L, we have r \cos \theta > d so r \cos \theta< 0 implies

        \[ |r \cos \theta + d | = - r \cos \theta - d \quad \implies \quad r = e (-r \cos \theta - d) \quad \implies \quad r = \frac{-ed}{1 + e \cos \theta}. \qquad \blacksquare \]

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