Eliminate the parameter and hence find the Cartesian equation of the curve. x=t + 1/t, y=t^2 + 1/t^2
Jackson94 Member Joined Jun 5, 2010 Messages 44 Gender Male HSC 2012 Nov 4, 2011 #1 Eliminate the parameter and hence find the Cartesian equation of the curve. x=t + 1/t, y=t^2 + 1/t^2
X xXnukerrrXx Member Joined Feb 17, 2011 Messages 51 Gender Male HSC N/A Nov 4, 2011 #2 (1) x=t+(1/t) (2) y=t^2+(1/t^2) similarly to the property a^2+b^2=(a+b)^2-2ab (1) x^2=(t+(1/t))^2=t^2+(1/t^2)+2 therefore to get t^2+(1/t^2) we minus the 2 x^2-2=t^2+(1/t^2) substitute t^2+(1/t^2) in equation (2) y=x^2-2 Last edited: Nov 4, 2011
(1) x=t+(1/t) (2) y=t^2+(1/t^2) similarly to the property a^2+b^2=(a+b)^2-2ab (1) x^2=(t+(1/t))^2=t^2+(1/t^2)+2 therefore to get t^2+(1/t^2) we minus the 2 x^2-2=t^2+(1/t^2) substitute t^2+(1/t^2) in equation (2) y=x^2-2
Jackson94 Member Joined Jun 5, 2010 Messages 44 Gender Male HSC 2012 Nov 4, 2011 #3 Thank you very much!