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Prove that the points of a conic section satisfy a given polar equation

Consider a conic section C with a horizontal directrix which is at a distance d above a focus at the origin. Prove that the points on C satisfy the polar equation

    \[ r = \frac{ed}{e \sin \theta + 1} \]

if the C is a ellipse or parabola and if C is a hyperbola then the points on the right branch satisfy the polar equation

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


Proof. From Theorem 13.17 (page 501 of Apostol) (taking F at the origin) we have

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

If we express X in polar coordinates, and take L to be horizontal, then we have N = \mathbf{j}. So, \lVert X \rVert = r, and X \cdot N = r \sin \theta. Therefore,

    \[ r = e | r \sin \theta - d|. \]

If X lies below the directrix L, then r \sin \theta < d, so | r \sin \theta - d| = d - r \sin \theta. Therefore,

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

If X lies above the directrix L, then r \sin \theta > d, so |r \sin \theta - d| = r  \sin \theta - d and we have

    \[ r = e (r \sin \theta - d) \quad \implies \quad r = \frac{ed}{e \sin \theta - 1}. \]

In this last case we also have e > 1 since r > 0 so this is a hyperbola. \qquad \blacksquare

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