Author Topic: Quadratics!  (Read 1322 times)

zara

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Quadratics!
« 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!

nid404

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Re: Quadratics!
« Reply #1 on: September 29, 2009, 08:07:03 am »
what's the answer?

Offline Ghost Of Highbury

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Re: Quadratics!
« Reply #2 on: September 29, 2009, 09:24:04 am »
i'm getting this answer..

k<8.06

is that correct??
divine intervention!

nid404

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Re: Quadratics!
« Reply #3 on: September 29, 2009, 09:29:04 am »
post your steps plz

Offline Ghost Of Highbury

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Re: Quadratics!
« Reply #4 on: September 29, 2009, 09:33:15 am »
lemme try again...i
ll post soon
divine intervention!

Offline astarmathsandphysics

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Re: Quadratics!
« Reply #5 on: September 29, 2009, 01:16:34 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)=0which simplifies to -4k^2+260=0 so k=\pm sqrt(65)
« Last Edit: September 29, 2009, 01:18:35 pm by astarmathsandphysics »

Offline Ghost Of Highbury

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Re: Quadratics!
« Reply #6 on: September 29, 2009, 01:19:06 pm »
yuppiee!!

my answer was correct...except the > sign
divine intervention!

Offline astarmathsandphysics

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Re: Quadratics!
« Reply #7 on: September 29, 2009, 01:19:54 pm »
I saw this yeasterday but forgot and only noticed it again this morning.

zara

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Re: Quadratics!
« Reply #8 on: September 29, 2009, 06:19:26 pm »
thanks astar n adi!
i figured out my mistake later...=]