IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: zara on September 28, 2009, 10:26:58 pm

Title: Quadratics!
Post by: zara on September 28, 2009, 10:26:58 pm: Find the value of k such tht the st. line y=2x+k meets the curve x²+2xy+2y²=5.

i started buh got stuck in between....

answer urgently anyone if possible!
thanks!
Title: Re: Quadratics!
Post by: nid404 on September 29, 2009, 08:07:03 am: what's the answer?
Title: Re: Quadratics!
Post by: Ghost Of Highbury on September 29, 2009, 09:24:04 am: i'm getting this answer..

k<8.06

is that correct??
Title: Re: Quadratics!
Post by: nid404 on September 29, 2009, 09:29:04 am: post your steps plz
Title: Re: Quadratics!
Post by: Ghost Of Highbury on September 29, 2009, 09:33:15 am: lemme try again...i
ll post soon
Title: Re: Quadratics!
Post by: astarmathsandphysics on September 29, 2009, 01:16:34 pm: Quote from: Z.J. on September 28, 2009, 10:26:58 pm
Find the value of k such tht the st. line y=2x+k meets the curve x²+2xy+2y²=5.

i started buh got stuck in between....

answer urgently anyone if possible!
thanks!

$x^2+2xy+y^2=5$

Substitute y=2x+k to get $x^2+2x(2x+k)+2(2x+k)^2=5$ so $x^2+4x^2+2xk+8x^2+8xk+2k^2-5=0$
so $13x^2+10kx+2k^2-5=0$

Since the curve meets the line at only one point the discriminant= $b^2-4ac=0$ where a=13, b=10k and c= $2k^2-5$

We get $(10k)^2-4*13*(2k^2-5)=0$ which simplifies to -4k^2+260=0 so k= $\pm sqrt(65)$
Title: Re: Quadratics!
Post by: Ghost Of Highbury on September 29, 2009, 01:19:06 pm: yuppiee!!

my answer was correct...except the > sign
Title: Re: Quadratics!
Post by: astarmathsandphysics on September 29, 2009, 01:19:54 pm: I saw this yeasterday but forgot and only noticed it again this morning.
Title: Re: Quadratics!
Post by: zara on September 29, 2009, 06:19:26 pm: thanks astar n adi!
i figured out my mistake later...=]