Preliminary Trigonometry (1 Viewer)

calebwhitaker199 · Jun 8, 2015

Could somebody please post a worked solution to this, the answer is
x^2 + y^2 +4y-5=0
Thank you

leehuan · Jun 8, 2015

This looks like parametrics more than trig but ok.

$From\quad x=3\sin { \theta } \\ { x }^{ 2 }=9\sin ^{ 2 }{ \theta } \\ 9\sin ^{ 2 }{ \theta } ={ x }^{ 2 }\quad -----------(1)\\ From\quad y=3\cos { \theta } -2\\ 3\cos { \theta } =y+2\\ Squaring:\\ 9\cos ^{ 2 }{ \theta } ={ y }^{ 2 }+4y+4\quad -------(2)\\ Add\quad (1)\quad and\quad (2):\\ 9\sin ^{ 2 }{ \theta } +9\cos ^{ 2 }{ \theta } ={ x }^{ 2 }+({ y }^{ 2 }+4y+4)\\ 9\left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) ={ x }^{ 2 }+{ y }^{ 2 }+4y+4\\ 9*1={ x }^{ 2 }+{ y }^{ 2 }+4y+4\\ { x }^{ 2 }+{ y }^{ 2 }+4y+4=9\\ { x }^{ 2 }+{ y }^{ 2 }+4y-5=0$

Preliminary Trigonometry (1 Viewer)

calebwhitaker199

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leehuan

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