Just a parametrics question (1 Viewer)

Aysce · Sep 30, 2011

Find the Cartesian equation:

[ a(p+q) , apq ] and pq =-4

I got y = -4a, but answer says y = -4

bleakarcher · Sep 30, 2011

probably a printing error

Aysce · Sep 30, 2011

Ugh how about this one, tried many times and failed -,-.

Find Cartesian equation of:

x= p+q

y= p^2 + q^2 Given that p^2 + q^2+ 6pq=0

taeyang · Sep 30, 2011

$y = p^{2}+q^{2} \\ y = (p+q)^2 -2pq...(1) \\\\ $however$\\\\p^2 + q^2 -6pq = 0\\ \and$ x^2 = (p+q)^2 \\\\ (p+q)^2 -8pq=0\\ x^2-8pq = 0\\ x^2 = 8pq\\ x^2/4 = 2pq ....(2)\\\\ \therefore $substitute (2) into (1)$\\\\ y = x^2 -\frac{x^2}{4}$

I am tired and so it could be wrong, and plus the working is not as good as it could have been. I just hope you can follow it ^.^

Aysce · Sep 30, 2011

I understand

but the answer you got to is apparently wrong, it's x^2 = 2/3 y :/ But thanks for trying to answer my question

Drongoski · Sep 30, 2011

I think taeyang made a simple slip somewhere.

$p^2+q^2+6pq = 0 \implies y+6pq=9 \implies y = -6pq$

$x^2 = (p+q)^2 = p^2 + q^2 + 2pq = p^2 + q^2 + 6pq - 4pq = -4pq = \frac {4}{6} y = \frac {2}{3}y$

AAEldar · Sep 30, 2011

$y = p^{2}+q^{2} \\ y = (p+q)^2 -2pq...(1) \\\\ $however$\\\\p^2 + q^2+6pq = 0\\ \and$ x^2 = (p+q)^2 \\\\ (p+q)^2+4pq=0\\ x^2+4pq = 0\\ x^2 = -4pq\\ \frac{x^2}{2} = -2pq ....(2)\\\\ \therefore $substitute (2) into (1)$\\\\ y = x^2 +\frac{x^2}{2} \\\\ y=\frac{3x^2}{2} \\\\ x^2=\frac{2y}{3}$

Was a slight error in taeyang's working.

taeyang · Oct 1, 2011

AAEldar said:
$y = p^{2}+q^{2} \\ y = (p+q)^2 -2pq...(1) \\\\ $however$\\\\p^2 + q^2+6pq = 0\\ \and$ x^2 = (p+q)^2 \\\\ (p+q)^2+4pq=0\\ x^2+4pq = 0\\ x^2 = -4pq\\ \frac{x^2}{2} = -2pq ....(2)\\\\ \therefore $substitute (2) into (1)$\\\\ y = x^2 +\frac{x^2}{2} \\\\ y=\frac{3x^2}{2} \\\\ x^2=\frac{2y}{3}$

Was a slight error in taeyang's working.

Drongoski said:
I think taeyang made a simple slip somewhere.

$p^2+q^2+6pq = 0 \implies y+6pq=9 \implies y = -6pq$

$x^2 = (p+q)^2 = p^2 + q^2 + 2pq = p^2 + q^2 + 6pq - 4pq = -4pq = \frac {4}{6} y = \frac {2}{3}y$

Oh man how embarrassing, sorry for that

Just a parametrics question (1 Viewer)

Aysce

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