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Author Topic: Help in maths p1!  (Read 1070 times)

Offline immortal

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Help in maths p1!
« on: April 20, 2010, 06:36:19 pm »
Find the values of k for which the line y=kx-2  meets da curve y2=4x-x2
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Offline past-papers

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Re: Help in maths p1!
« Reply #1 on: April 20, 2010, 06:57:15 pm »
simultaneous equations:

y^2 => k^2x^2 -4kx +4 = 4x - x^2
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nid404

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Re: Help in maths p1!
« Reply #2 on: April 20, 2010, 06:58:27 pm »
Quote from: immortal on April 20, 2010, 06:36:19 pm
Find the values of k for which the line y=kx-2  meets da curve y2=4x-x2

Post the paper up
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Offline past-papers

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Re: Help in maths p1!
« Reply #3 on: April 20, 2010, 07:00:04 pm »
wait, actually you're missing something --> post the whole question
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Offline immortal

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Re: Help in maths p1!
« Reply #4 on: April 20, 2010, 07:13:46 pm »
I posted da whole question which i got 4 ma Mocks.da only thing i missed out waz dat it waz a [4] marks question.
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Offline student

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Re: Help in maths p1!
« Reply #5 on: April 20, 2010, 07:50:16 pm »
I didnt do p1, but..

y2=4x-x2 is a circle, isnt it?
With 0=<x=<4.

And y=kx-2 would pass through this circle at k>0 to the infinity.
« Last Edit: April 20, 2010, 07:58:28 pm by student »
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Offline tmisterr

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Re: Help in maths p1!
« Reply #6 on: April 21, 2010, 03:58:52 pm »
Quote from: past-papers on April 20, 2010, 06:57:15 pm
simultaneous equations:

y^2 => k^2x^2 -4kx +4 = 4x - x^2

You probably are missing something but if not continuing from past papers equation

k2x2-x2-4kx-4x+4=0 so as you know if it meets the curve it must either have a repeated root (b2=4ac or it b2>4ac

so (-4k-4)2>or=4*(k2+1)*4
solving this equation gives you k>or=0.
« Last Edit: April 21, 2010, 04:04:11 pm by tmisterr »
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nid404

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Re: Help in maths p1!
« Reply #7 on: April 21, 2010, 04:32:12 pm »
yup that what i did and concluded there is something missing :-\
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Offline immortal

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Re: Help in maths p1!
« Reply #8 on: April 21, 2010, 05:27:28 pm »
y=kx-2  meets da curve y2=4x-x2, so it has a root.

(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2-(4k+4)x+4=0

as da eqn haz 1 root b2-4ac=0
(4k+4)2-(4*(k2+1)*4)=0
16k2+32k+16-(16k2+16)=0

but i get 32k=0 which is probably wrong!
any ideas Guyz ???


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Offline student

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Re: Help in maths p1!
« Reply #9 on: April 21, 2010, 05:34:13 pm »
Quote
Find the values of k for which the line y=kx-2  meets da curve y2=4x-x2

There's not one root.
And y=kx-2 does meet y2=4x-x2 at any values of k>0 or equal to 0..
Otherwise there might be some special 'p1 method' for it?
« Last Edit: April 21, 2010, 05:41:00 pm by student »
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