convert r= (6/2-3 sin @) to a cartesian equation
S she18 New member Joined Nov 9, 2005 Messages 12 Nov 17, 2005 #1 convert r= (6/2-3 sin @) to a cartesian equation
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Nov 17, 2005 #2 Six divided by two is three... or did you mean "r = 6/(2 - 3sin(@))"? Thank you. Eliz.
S she18 New member Joined Nov 9, 2005 Messages 12 Nov 17, 2005 #3 ya wHAT YOU wrote out is what i meant
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Nov 17, 2005 #4 Since y = rsin(@), I would start by cross-multiplying and isolating "rsin(@)". Eliz.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Nov 18, 2005 #5 Hello, she18! Convert .r = 62 − 3⋅sinθ\displaystyle r\:=\:\frac{6}{2\,-\,3\cdot\sin\theta}r=2−3⋅sinθ6 .to a cartesian equation Click to expand... We have: .r(2 − 3⋅sinθ) = 6 ⇒ 2r − 3⋅rsinθ = 6 ⇒ 2r = 3⋅rsinθ + 6\displaystyle r(2\,-\,3\cdot\sin\theta)\:=\:6\;\;\Rightarrow\;\;2r\,-\,3\cdot r\sin\theta\:=\:6\;\;\Rightarrow\;\;2r\:=\;3\cdot r\sin\theta\,+\,6r(2−3⋅sinθ)=6⇒2r−3⋅rsinθ=6⇒2r=3⋅rsinθ+6 Convert: .2x2 + y2 = 3y + 6\displaystyle 2\sqrt{x^2\,+\,y^2}\:=\:3y\,+\,62x2+y2=3y+6 Square: .\(\displaystyle 4(x^2\,+\,y^2)\:=\3y\,+\,6)^2\;\;\Rightarrow\;\;4x^2\,+\,4y^2\:=\:9y^2\,+\,36y\,+\,36\) Therefore: .5x2 − 5y2 − 36y = 36\displaystyle 5x^2\,-\,5y^2\,-\,36y\:=\:365x2−5y2−36y=36 . . . hyperbola
Hello, she18! Convert .r = 62 − 3⋅sinθ\displaystyle r\:=\:\frac{6}{2\,-\,3\cdot\sin\theta}r=2−3⋅sinθ6 .to a cartesian equation Click to expand... We have: .r(2 − 3⋅sinθ) = 6 ⇒ 2r − 3⋅rsinθ = 6 ⇒ 2r = 3⋅rsinθ + 6\displaystyle r(2\,-\,3\cdot\sin\theta)\:=\:6\;\;\Rightarrow\;\;2r\,-\,3\cdot r\sin\theta\:=\:6\;\;\Rightarrow\;\;2r\:=\;3\cdot r\sin\theta\,+\,6r(2−3⋅sinθ)=6⇒2r−3⋅rsinθ=6⇒2r=3⋅rsinθ+6 Convert: .2x2 + y2 = 3y + 6\displaystyle 2\sqrt{x^2\,+\,y^2}\:=\:3y\,+\,62x2+y2=3y+6 Square: .\(\displaystyle 4(x^2\,+\,y^2)\:=\3y\,+\,6)^2\;\;\Rightarrow\;\;4x^2\,+\,4y^2\:=\:9y^2\,+\,36y\,+\,36\) Therefore: .5x2 − 5y2 − 36y = 36\displaystyle 5x^2\,-\,5y^2\,-\,36y\:=\:365x2−5y2−36y=36 . . . hyperbola
3y\,+\,6)^2\;\;\Rightarrow\;\;4x^2\,+\,4y^2\:=\:9y^2\,+\,36y\,+\,36\)