satellite73
  • satellite73
how do we show, without a picture, that the ellipse \[\frac{9+1)^2}{25}+\frac{9y+2)^2}{4}=1\] does not intersect the parabola \[y=(x+1)^2+1\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
damn typo!
anonymous
  • anonymous
\[\frac{(x+1)^2}{25}+\frac{(y+2)^2}{4}=1\]
anonymous
  • anonymous
my attempt \[y=(x+1)^2+1\\ y-1=(x+1)^2\] so \[\frac{y-1}{25}+\frac{(y+2)^2}{4}=1\]

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More answers

anonymous
  • anonymous
then i multiplied by 100 and get \[4(y-1)+25(y+2)^2=100\] but that has a solutions!
Mr_Perfection_xD
  • Mr_Perfection_xD
i wish i was a mathlete
anonymous
  • anonymous
lol me too!
pooja195
  • pooja195
.-. well im one and this makes no sense :P
rsadhvika
  • rsadhvika
\(y = (x+1)^2+1 \ge 1\) \(y = -2\pm 2\sqrt{1-\frac{(x+1)^2}{25}} \le -2 \pm 2 = 0\)
Mr_Perfection_xD
  • Mr_Perfection_xD
i wanna go beyond the call of duty
geerky42
  • geerky42
Maybe somehow check the range of these functions and then we can show that intersection of these ranges is empty set.
geerky42
  • geerky42
like what I think @rsadhvika is trying to do.
anonymous
  • anonymous
yeah i graphed and it was obvious but i was looking for algebra i guess the range is the way to go
anonymous
  • anonymous
probably just should have found the vertices and have been done with it at that point thanks !
jim_thompson5910
  • jim_thompson5910
geerky42 is right on the money solving \(4(y-1)+25(y+2)^2=100\) gets you these approximate solutions \(\Large y \approx 0.0381\) \(\Large y \approx -4.1981\) but neither of these y values are in the range of \(\Large y = (x+1)^2 + 1\)
anonymous
  • anonymous
yeah minor axis has length 2 and the center is \((-1,-2)\) so the whatever they are called are \((-1,0)\) and \((-1,-4)\)

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