Irregular Cone Geometry Problem: Finding the Area of an Elliptical Base

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To find the area of the base of an elliptical cone using the length from the center to the vertex and the angle of the sides, one can utilize the formula A=πab, where a and b are the semi-major and semi-minor axes. The discussion suggests employing trigonometry to derive these axes by dropping a perpendicular from the vertex to the base, creating a right triangle. A helpful approach involves slicing the cone to obtain a circular cross-section, which can simplify the calculations. Despite initial challenges in geometry guides, applying these methods may lead to a solution. Understanding the relationship between the cone's dimensions and the elliptical base is crucial for solving the problem.
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Homework Statement



I'm wondering whether or not it is possible to get the area of the base of an elliptical, non-right cone if the following two parameters are known:

- length from center of ellipse up to vertex
- angle that the sides make

Here is a simple visual: http://i.imgur.com/M6jT5.png

Homework Equations



A=\pi ab

The Attempt at a Solution



I've tried looking at geometry guides with no success. I'm thinking I might be able to use some clever trig and drop a right angle down from the vertex in order to get the length of the semi-minor/major axis, but I can't reason it out. any advice would be appreciated.
 
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Hi Appity! :smile:

Hint: slice the cone so that you get a circular cross-section, then use trig to find the area of the projection. :wink:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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