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Additional Math Help HERE ONLY...!

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BlackBunny103:
Hey guys, would you please help me to solve this question, it's from MJ 2003 paper 1.
Thanxx :D

slvri:
y=kx-2
y2=4x-x2
substitute the first equation in the second
(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2+(-4k-4)x+4=0
now the line meets the curve provided that for the above equation the discriminant or b2-4ac>=0
where a=k2+1,b=-4k-4 and c=4
b2-4ac>=0
(-4k-4)2-4(k2+1)(4)>=0
16k2+32k+16-16k2-16>=0
32k>=0
k>=0
i hope this helped, and is this the correct answer? :)

BlackBunny103:

--- Quote from: slvri on June 07, 2010, 09:32:17 am ---y=kx-2
y2=4x-x2
substitute the first equation in the second
(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2+(-4k-4)x+4=0
now the line meets the curve provided that for the above equation the discriminant or b2-4ac>=0
where a=k2+1,b=-4k-4 and c=4
b2-4ac>=0
(-4k-4)2-4(k2+1)(4)>=0
16k2+32k+16-16k2-16>=0
32k>=0
k>=0
i hope this helped, and is this the correct answer? :)

--- End quote ---

Oh great, thanxx +rep. Yes this is the correct answer ;)
By the way are you taking Add Maths this year?

BlackBunny103:
Can you guys please help me with Q8 of MJ 2004 P1?
Thanxx :D

astarmathsandphysics:
here

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