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Compute the distance from focus to directrix for a conic section

Consider a conic section with focus at the origin, eccentricity e and directrix given by

    \[ e = \frac{1}{2}; \quad \text{directrix}: 3x + 4y = 25. \]

Compute the distance d from the focus to the directrix and find a polar equation for C. For a hyperbola, give a polar equation for each branch.


First, the distance from any point X to the directrix L is given by the equation

    \[ d(X,L) = \frac{|(X-P) \cdot N|}{\lVert N \rVert} \]

where P is any point on the directrix L. In this case we can choose P = \left( \frac{25}{3}, 0\right) as a point on the directrix and let X = (0,0) be the focus. Then the distance from the focus to the directrix is

    \[ d = \frac{P \cdot N}{\lVert N \rVert}. \]

Since a normal to the directrix is N = (3,4) (from the Cartesian equation for the directrix) we have

    \[ d = \frac{ \left( \frac{25}{3}, 0 \right) \cdot (3,4) }{5} = 5. \]

To obtain the polar equation for the conic section, since the focus is at the origin, we have the conic section is the set of points X such that

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

Letting X have polar coordinates r and \theta, (i.e., X = (r \cos \theta, r \sin \theta)), and letting N = \left( \frac{3}{5}, \frac{4}{5} \right) be the unit normal we have the polar equation

    \begin{align*}  && \lVert X \rVert &= e | X \cdot N - d | \\[9pt]  \implies && r &= \frac{1}{2} \left| \frac{3}{5} r \cos \theta + \frac{4}{5} r \sin \theta - 5\right| \\[9pt]  \implies && 2r &= 5 - \frac{3}{5} r \cos \theta - \frac{4}{5} r \sin \theta \\[9pt]  \implies && 10r &= 25 - 3 r \cos \theta - 4 r \sin \theta \\[9pt]  \implies && r &= \frac{25}{10 + 3 \cos \theta + 4 \sin \theta}. \end{align*}

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