Consider a conic section with focus at the origin, eccentricity 
and directrix given by
![\[ e = \frac{1}{2}; \quad \text{directrix}: 3x + 4y = 25. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fdfe4773490a20682166d77ad8869b58_l3.png "Rendered by QuickLaTeX.com")
Compute the distance 
from the focus to the directrix and find a polar equation for 
. For a hyperbola, give a polar equation for each branch.
First, the distance from any point 
to the directrix 
is given by the equation
![\[ d(X,L) = \frac{|(X-P) \cdot N|}{\lVert N \rVert} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-949a620d7618093c55e39ade2adaafb2_l3.png "Rendered by QuickLaTeX.com")
where 
is any point on the directrix . In this case we can choose 
as a point on the directrix and let 
be the focus. Then the distance from the focus to the directrix is
![\[ d = \frac{P \cdot N}{\lVert N \rVert}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8f1434654be5a6a2faa2db458c0ee083_l3.png "Rendered by QuickLaTeX.com")
Since a normal to the directrix is 
(from the Cartesian equation for the directrix) we have
![\[ d = \frac{ \left( \frac{25}{3}, 0 \right) \cdot (3,4) }{5} = 5. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-902c5b4c416ef9908d3afdfbb981352b_l3.png "Rendered by QuickLaTeX.com")
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 such that
![\[ \lVert X \rVert = e| X \cdot N - d|. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d8543785974efa686f6fb527211a7d7b_l3.png "Rendered by QuickLaTeX.com")
Letting have polar coordinates 
and 
, (i.e., 
), and letting 
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7aa1cff898e1f13983cea668c0e17763_l3.png "Rendered by QuickLaTeX.com")
Is there any way to know which way to calculate the absolute value (ie to take the positive or negative of whatever’s inside the absolute value)? The only thing I can think of is to calculate both versions and see which one implies r < 0
Thanks!