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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 \[y=(x+1)^2+1\]

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  1. anonymous
    • one year ago
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    damn typo!

  2. anonymous
    • one year ago
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    \[\frac{(x+1)^2}{25}+\frac{(y+2)^2}{4}=1\]

  3. anonymous
    • one year ago
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    my attempt \[y=(x+1)^2+1\\ y-1=(x+1)^2\] so \[\frac{y-1}{25}+\frac{(y+2)^2}{4}=1\]

  4. anonymous
    • one year ago
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    then i multiplied by 100 and get \[4(y-1)+25(y+2)^2=100\] but that has a solutions!

  5. Mr_Perfection_xD
    • one year ago
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    i wish i was a mathlete

  6. anonymous
    • one year ago
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    lol me too!

  7. pooja195
    • one year ago
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    .-. well im one and this makes no sense :P

  8. rsadhvika
    • one year ago
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    \(y = (x+1)^2+1 \ge 1\) \(y = -2\pm 2\sqrt{1-\frac{(x+1)^2}{25}} \le -2 \pm 2 = 0\)

  9. Mr_Perfection_xD
    • one year ago
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    i wanna go beyond the call of duty

  10. geerky42
    • one year ago
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    Maybe somehow check the range of these functions and then we can show that intersection of these ranges is empty set.

  11. geerky42
    • one year ago
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    like what I think @rsadhvika is trying to do.

  12. anonymous
    • one year ago
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    yeah i graphed and it was obvious but i was looking for algebra i guess the range is the way to go

  13. anonymous
    • one year ago
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    probably just should have found the vertices and have been done with it at that point thanks !

  14. jim_thompson5910
    • one year ago
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    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 \(\Large y = (x+1)^2 + 1\)

  15. anonymous
    • one year ago
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    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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