How can the vertices of an ellipse be found using polar equations?

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SUMMARY

The vertices of the ellipse defined by the polar equation r = 15/[3-2cos(θ)] can be determined using its symmetry and specific angle values. The equation can be rearranged to 5/[1-(2/3)cos(θ)], confirming the ellipse's eccentricity as e=2/3. The vertices are located at the points (15,0) and (3,π) in polar coordinates. To find these vertices, one can evaluate the equation at key angles, leveraging the symmetry about the x-axis.

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i_love_science
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Homework Statement
Conic: r = 15/[3-2cos(theta)]
Relevant Equations
conic general form when vertical directrix to left of pole:
r = ed/[1-e*cos(theta)], where d is the distance between focus at pole and the directrix
The text says that the following conic, r = 15/[3-2cos(theta)], can be rearranged to 5/[1-(2/3)cos(theta)]. The graph of the conic is an ellipse with e=2/3.
Then it says that the vertices lie at (15,0) and (3,pi). How did they find the vertices? Thanks.
 
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hello I_love_science. Here is a graph of your polar equation for reference.

Screen Shot 2021-01-20 at 11.56.03 PM.png


If you notice the ellipse is symmetric about an axis, you can plug in certain values of ##θ## to determine the vertices in polar coordinates ##(r, θ)## and convert them to ##(x, y)## coordinates. In this example, the ellipse is symmetric about the x axis, which makes finding vertices more streamlined.

If the ellipse was not symmetric about the x or y axis, then my first step would be to find the axis of symmetry.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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