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satellite73
 one year ago
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
\[y=(x+1)^2+1\]
satellite73
 one year ago
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 \[y=(x+1)^2+1\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{(x+1)^2}{25}+\frac{(y+2)^2}{4}=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my attempt \[y=(x+1)^2+1\\ y1=(x+1)^2\] so \[\frac{y1}{25}+\frac{(y+2)^2}{4}=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then i multiplied by 100 and get \[4(y1)+25(y+2)^2=100\] but that has a solutions!

Mr_Perfection_xD
 one year ago
Best ResponseYou've already chosen the best response.0i wish i was a mathlete

pooja195
 one year ago
Best ResponseYou've already chosen the best response.0.. well im one and this makes no sense :P

rsadhvika
 one year ago
Best ResponseYou've already chosen the best response.1\(y = (x+1)^2+1 \ge 1\) \(y = 2\pm 2\sqrt{1\frac{(x+1)^2}{25}} \le 2 \pm 2 = 0\)

Mr_Perfection_xD
 one year ago
Best ResponseYou've already chosen the best response.0i wanna go beyond the call of duty

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Maybe somehow check the range of these functions and then we can show that intersection of these ranges is empty set.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0like what I think @rsadhvika is trying to do.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i graphed and it was obvious but i was looking for algebra i guess the range is the way to go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0probably just should have found the vertices and have been done with it at that point thanks !

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0geerky42 is right on the money solving \(4(y1)+25(y+2)^2=100\) 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah 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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