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eliminate the parabola and hence find the cartesian equation of the curve with x=1+2tantheta and y=3sectheta-4
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"...and hence..." I seem to remember that redundant statement being in JonBenet Ramsey's hostage letter. This might help: http://community.boredofstudies.org/12/mathematics/289228/locus-parabola-help.html
x=1+2tantheta tantheta =(x-1)/2 sec^2theta=1+tan^2(theta) =1+((x-1)/2)^2 y=3sectheta-4 3sectheta=y+4 square on both sides 9sec^2theta=(y+4)^2 9(1+((x-1)/2)^2)=(y+4)^2 9=(y+4)^2-((x-1)^2/4 (y+4)^2/9-((x-1)^2/36 =1 Source: http://www.mathskey.com/question2answer/
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