IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: halosh92 on February 21, 2010, 01:10:40 pm
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plz can anyone solve this!!!!!!!!!!!!!! my head broke!!!!!!!!!!!
the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
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Can u tell which paper this is from...cuz something seems wrong...
the middle term in AP2 should be 12t and not -24t...
if i knw the paper i can cross check
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Can u tell which paper this is from...cuz something seems wrong...
the middle term in AP2 should be 12t and not -24t...
if i knw the paper i can cross check
it is a pastpaper question but they put in our C2 skool textbook the year is not written but i think its pretty old.
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Is f the question correct?
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Is f the question correct?
i rili have no idea ???
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ok...do u have ans at the back of your book?
If yes...then i can proceed my way and check if the other ans are right
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Should be home in 2 hours.
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ppl, this question is a screwed one.
A point cannot have 3 co-ordinates. hehe
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ppl, this question is a screwed one.
A point cannot have 3 co-ordinates. hehe
well spotted :o
This is completely nuts :P
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well spotted :o
This is completely nuts :P
i totally agree ! i cant believe wat kind of people sit and make up such questions
yea i have the answers but there is nothing for the a) part :
b) t=2
c) square root 101 = 10.0
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oh yah i saw that question in the text book...didn't know how to solve so i skipped
but a point can have 3 coordinate, if you're considering the z axis, but then again we haven't been taught into such details
im sure its something simple, that we're just not getting it.
anyways i have one more question
could someone plz solve C2 question 8 Jun05,
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assuming they are vectors A & P and using their first eqn...which i also think is wrong
U can solve the quadratic...
x=(-b +/- root of (b2-4ac))/ 2a
u get t as 2
AP2= 6(4)-24(2)+125=101
AP=root of 101= 10..
strange...
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the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
a)AP=sqrt((t-8)^2+(t--6)^2+(5-2t)^)
=sqrt(t^2-16t+64+t^2+12t+36+25-20t+4t^2)=sqrt(6t^2-24t+125)
b)AP^2=6t^2-24t+125
diffferentiate and put=0 and solve
12t-24=0 so t=2
c)AP^2 =6*2^2-24*2+125=24-48+125=101
least distance=sqrt(101)
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Which board is this? Not cie?
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the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
a)AP=sqrt((t-8)^2+(t--6)^2+(5-2t)^)
=sqrt(t^2-16t+64+t^2+12t+36+25-20t+4t^2)=sqrt(6t^2-24t+125)
b)AP^2=6t^2-24t+125
diffferentiate and put=0 and solve
12t-24=0 so t=2
c)AP^2 =6*2^2-24*2+125=24-48+125=101
least distance=sqrt(101)
thanks alooot nid and astar ;D ;D ;D
yea its is cie board!
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thanks alooot nid and astar ;D ;D ;D
yea its is cie board!
its edexcel not cie...
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oh yah i saw that question in the text book...didn't know how to solve so i skipped
but a point can have 3 coordinate, if you're considering the z axis, but then again we haven't been taught into such details
im sure its something simple, that we're just not getting it.
anyways i have one more question
could someone plz solve C2 question 8 Jun05,
8a) x^2+y^2-10x+9=0
x^2-10x+y^2+9=0
(x-5)^2-25+y^2+9=0
(x-5)^2+y^2+9=0
hence
(x-5)^2+y^2-16=0
center: (5,0)
8b.) (x-5)^2+y^2=16
square root 16 =4=radius
8c) radius= 4 center : (5,0)
substitute y in the f(x) equation with 0
therefore:
x^2-10x+9=0
(x-9)(x-1)=0
x=9
x=1
hence
(9,0) and (1,0)
8d) gradiant of AT
m= -2/7
y-y=m(x-x)
y-0=-2/7(x-5)
hence
y=-2/7(x-5)