Qualification > Math

Help in maths p1!

<< < (2/2)

student:
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.

tmisterr:

--- Quote from: past-papers on April 20, 2010, 06:57:15 pm ---simultaneous equations:



--- End quote ---

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.

nid404:
yup that what i did and concluded there is something missing :-\

immortal:
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 ???

student:

--- Quote ---Find the values of k for which the line y=kx-2  meets da curve y2=4x-x2
--- End quote ---

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?

Navigation

[0] Message Index

[*] Previous page