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Mathematics 6 Online
OpenStudy (roberts.spurs19):

Please help! Find the Cartesian form of

OpenStudy (roberts.spurs19):

Thank you!

OpenStudy (roberts.spurs19):

\[r(1 + \sin \theta) = a\]

OpenStudy (misty1212):

you are welcome dear

OpenStudy (misty1212):

is \(a\) a constant?

OpenStudy (roberts.spurs19):

yes, I think so

OpenStudy (misty1212):

if you multiply you can get \[r+r\sin(\theta)=a\] and since \\[r\sin(\theta)=y\] you can rewrite as \[r+y=a\]

OpenStudy (misty1212):

after that i think the best you can do is to rewrite \[r=\sqrt{x^2+y^2}\] gets you \[\sqrt{x^2+y^2}+y=a\] then maybe some algebra to get rid of the radical might be an easier way, but i am not sure

hartnn (hartnn):

yeah, \(\sqrt{x^2+y^2} = a-y\) squaring that would look nice :)

OpenStudy (roberts.spurs19):

Thank you so much! :)

OpenStudy (misty1212):

you are welcome dear

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