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Find a polar equation for a given conic section

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

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

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 = (1,0) 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 = (1,1) (from the Cartesian equation for the directrix) we have

    \[ d = \frac{ (1,0) \cdot (1,1) }{\sqrt{2}} = \frac{\sqrt{2}}{2}. \]

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{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) be the unit normal we have the two polar equations. The first is given by

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

For the other polar equation (just reversing the sign on the absolute value above) we have

    \[ r = \frac{1}{\cos \theta + \sin \theta - \frac{\sqrt{2}}{2}}. \]

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