Rotating curves [40 marks]...Title Untitled Created Date 11/9/2020 5:14:01 AM
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Copyright Andrew Chambers 2020. Not licensed for commercial use. Visit ibmathsresources.com for over 300 exploration ideas, exploration guides and beautiful mathematics! 3 Rotating curves [40 marks] 1. [Maximum marks: 15] (a) !" !" = !"#$% !"#$% = !"#$ (M1) ! = (!"#$)! A1 [2 marks] (b) For ! ! : ! = 1 !"# ! ! − 1 !"# ! ! = !!! ! ! ! = 1 !"# ! ! + 1 !"# ! ! = !! ! ! (M1) shown for substitution into one pair of equations !!! ! ! , !! ! ! A1 For ! ! : (0, 2 ) A1 For ! ! : −1,1 A1 [4 marks]
Transcript of Rotating curves [40 marks]...Title Untitled Created Date 11/9/2020 5:14:01 AM
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3Rotatingcurves[40marks]
1.[Maximummarks:15](a)!"!" = !"#$%!"#$% = !"#$(M1)! = (!"#$)!A1[2marks](b)For!! :! = 1 !"# !! − 1 !"# !
! = !!! !!
! = 1 !"# !
! + 1 !"# !! =!! !! (M1) shown for substitution into one pair of
equations!!! !
! , !! !! A1
For!!:(0, 2)A1For!!:−1,1 A1[4marks]
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4(c)
M1A1Allthepointslieonacirclecentred(0,0)radiusroot2.A1A1[4marks](d)! = !"#$% − !"#$% (1)! = !"#$% + !"#$% (2)!! = !!!!"!! + !!!"#!! − 2!"#$%&%'(&(3)(M1)!! = !!!"#!! + !!!"#!! + 2!"#$!"!#$"(4)(M1)!! + !! = (!! + !!)(!"#!! + !"#!!)(3)+(4)(M1)!! + !! = (!! + !!)A1Circlecentred(0,0),radius (!! + !!)A1[5marks]
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52.[Maximummarks:25](a)
Useofasketchabove–alternateanglemarkedastheta(A1),angleaboveaspi/2–theta(A1)[2marks](b)!" = !"#$%!" = ! ! !"#$A1forbothThexcoordinateofPisgivenbyAB–PD.M1Thereforexcoordinategivenby!"#$% − ! ! !"#$AG!" = !"#$%!" = ! ! !"#$A1forbothTheycoordinateofPisgivenbyBC+PD.M1Thereforeycoordinategivenby!"#$% + ! ! !"#$ AG[4marks]
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6(c)! = !"#$ ! − ! ! !"#$ (1)! = !"#$ ! + ! ! !"#$ (2)!"#$% = !!"#! ! − ! ! !"#$%&!$ (3)!"#$% = !!"#! ! + ! ! !"#$#%&$ (4)M1attempttosquarebothsides. A1bothcorrectlydone.!"#$% + !"#$% = !(1)(3)+(4)(M1)Substituteinto(2)(equivalentmethodforsubinto(1))! = (!"#$% + !"#$%)!"# ! + ! !"#$% + !"#$% !"#$ (M1)! 1− !"#!! = !"#$%$&'% + ! !"#$% + !"#$% !"#$ (M1)! !"#!! = !"#$%$&'% + ! !"#$% + !"#$% !"#$! !"#$ = !"#$% + ! !"#$% + !!"#$ (M1)! !"#$ − !"#$% = ! !"#$% + !"#$% AG[6marks](d)! ! = 2! + 3! !"#$% + !"#$% = 2 !"#$% + !"#$% + 3(M1)! = !!!"#$% − !"#$% = !(!"#$% + !"#$%)!"#$ !! − !"#$
!! = 2 !"#$ !
! + !"#!!! + 3(M1)(A1)
! !
! − !!! = 2 ! !
! + !!! + 3(M1)
! − ! = 2 ! + ! + 3 !
!(M1)
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7! = −3! − 3 2AG[5marks](e)3! + 1! + 3 2 = 0−2! + 1! − 3 = 0(M1)Thentheanglebetweenthelinesisgivenby:!"#$ = (!)(!)!(!)(!!)
! !! !(!)(!) (M1)!"#$ = −1 !"#$ = 1A1! = !!AG[3marks](f)! ! = !!! !"#$% + !"#$% = !"#$% + !"#$% !(M1)! = !!!"#$% − !"#$% = !(!"#$% + !"#$%)!"#$ !! − !"#$
!! = !"#$ !
! + !"#$!! !(M1)
! !
! − !!! = ! !
! + !!!
!
! !
! − !!! =
!!! +
!!! + !"(A1)
0 = !!
! + !!! + !!! − !
!! + !"(A2)
[5marks]
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