Interesting geometric problem

Interesting geometric problem

Postby Guest » Wed Jan 16, 2019 6:33 pm

Hi All,

To better describe the problem I have uploaded here:

https://docs.google.com/document/d/13l9MZaSrdoh7oF7m75OB9jTRPnXlL_e7VXjUCFkqQBc/edit?usp=sharing

Feel free to ask any questions.

Any help is appreciated!
Guest
 

Re: Interesting geometric problem

Postby Guest » Wed Jan 16, 2019 6:35 pm

Sorry I realized I have posted this 'Fractions' after the fact.
Guest
 

Re: Interesting geometric problem

Postby Baltuilhe » Wed Jan 16, 2019 9:01 pm

Good night!

Not so simple... but I think I've solved :)

Ellipse equation:
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]

Data:
[tex]a=\dfrac{Aw\cdot\left(1-P\right)}{2}[/tex]
[tex]b=\dfrac{Ah\cdot\left(1-P\right)}{2}[/tex]

Now you have a and b.

Given Bwa and Bha:

[tex]x=\dfrac{Bwa}{Bha}\cdot y[/tex]

Because of that, using this x in the Ellipse equation, we have y:
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]

Using x:
[tex]\dfrac{\left(\dfrac{Bwa}{Bha}\cdot y\right)^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]

Arranging:
[tex]\dfrac{\dfrac{Bwa^2}{Bha^2}}{a^2}\cdot y^2+\dfrac{y^2}{b^2}=1[/tex]

Preparing to isolate y:
[tex]y^2\cdot\left(\dfrac{\dfrac{Bwa^2}{Bha^2}}{a^2}+\dfrac{1}{b^2}\right)=1[/tex]

Isolate y:
[tex]y=\dfrac{1}{\sqrt{\dfrac{\dfrac{Bwa^2}{Bha^2}}{a^2}+\dfrac{1}{b^2}}}[/tex]

Now return to x:
[tex]x=\dfrac{Bwa}{Bha}\cdot\dfrac{1}{\sqrt{\dfrac{\dfrac{Bwa^2}{Bha^2}}{a^2}+\dfrac{1}{b^2}}}[/tex]

Finally:
[tex]Bwb=2x\\\\Bhb=2y[/tex]

I hope I have helped!

Baltuilhe
 
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Re: Interesting geometric problem

Postby Guest » Thu Jan 17, 2019 11:51 am

At a glance looks all good! Fantastic. Thank you so much.
Guest
 


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