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1ace Group Title

Find the points on the ellipse 100x^2+y^2=100 that are farthest away from the point (1,0)

  • 2 years ago
  • 2 years ago

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  1. freckles Group Title
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    You know the distance formula?

    • 2 years ago
  2. freckles Group Title
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    You want to maximize that using the points (1,0) and (x,y)

    • 2 years ago
  3. 1ace Group Title
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    ok i know y^2=(100-100x^2)

    • 2 years ago
  4. 1ace Group Title
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    so d^2=x^2 + (100-100x^2)^2

    • 2 years ago
  5. 1ace Group Title
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    but im not sure about the x^2 part is it (1-x)^2 because the point is (1,0)?

    • 2 years ago
  6. 1ace Group Title
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    or am I just doing this all wrong? :\

    • 2 years ago
  7. freckles Group Title
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    \[d=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{((x-1)^2+(y-0)^2}=\sqrt{(x-1)^2+y^2}\] I used the points given (1,0) and (x,y)

    • 2 years ago
  8. freckles Group Title
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    Okay and you squared the distance! Good job! :)

    • 2 years ago
  9. freckles Group Title
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    \[d^2=(x-1)^2+y^2\] Replace y^2 with what you found :)

    • 2 years ago
  10. 1ace Group Title
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    yep i did that ! :o

    • 2 years ago
  11. freckles Group Title
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    \[100x^2+y^2=100 => y^2=100-100x^2\]

    • 2 years ago
  12. freckles Group Title
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    Why did you square y^2?

    • 2 years ago
  13. freckles Group Title
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    \[d^2=(x-1)^2+100-100x^2\]

    • 2 years ago
  14. 1ace Group Title
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    OOOHH omg hahaahah thats where i went wrong?!

    • 2 years ago
  15. 1ace Group Title
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    so then after i take the derivative is do i find the roots ? -199x-1=0? that doesnt seem right :s

    • 2 years ago
  16. freckles Group Title
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    Nope nope.... So when you found ((x-1)^2)' You used chain rule right?

    • 2 years ago
  17. freckles Group Title
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    We know that (100)'=0

    • 2 years ago
  18. freckles Group Title
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    (-100x^2)'=-100(x^2)' we know to use power rule here

    • 2 years ago
  19. 1ace Group Title
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    yes so i get 2(x-1)*1 + 0 -200x

    • 2 years ago
  20. freckles Group Title
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    You need to distribute that 2. 2(x-1)=2x-1(2)=2x-2

    • 2 years ago
  21. 1ace Group Title
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    okay!

    • 2 years ago
  22. freckles Group Title
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    Right!: )

    • 2 years ago
  23. freckles Group Title
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    \[\text{ Set } (d^2)'=0\]

    • 2 years ago
  24. freckles Group Title
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    \[2x-2-200x=0\]

    • 2 years ago
  25. 1ace Group Title
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    ok but dont i have to divide by the 2d from when i differentiate (d^2)'

    • 2 years ago
  26. freckles Group Title
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    No no.

    • 2 years ago
  27. 1ace Group Title
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    why not ?

    • 2 years ago
  28. freckles Group Title
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    Minimizing d^2 will give us the same results as minimizing d

    • 2 years ago
  29. 1ace Group Title
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    ok, so i get x=1/98 ?

    • 2 years ago
  30. 1ace Group Title
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    then would that be my point? x= 1/98 and subbing that into the original equation y= 9.99948?

    • 2 years ago
  31. 1ace Group Title
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    also i only get 1 answer, shouldnt there be another point also?

    • 2 years ago
  32. freckles Group Title
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    \[y=\pm \sqrt{100-100x^2} \text{ correct?}\]

    • 2 years ago
  33. freckles Group Title
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    Say yes. lol

    • 2 years ago
  34. 1ace Group Title
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    but my x is only 1 point? and my y's will be 2 points? sorry im just a little confused

    • 2 years ago
  35. freckles Group Title
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    \[(\frac{1}{99}, \sqrt{100-100(\frac{1}{99})^2}) \text{ and } (\frac{1}{99},-\sqrt{100-100(\frac{1}{99})^2})\]

    • 2 years ago
  36. 1ace Group Title
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    okay but why 1/99? i thought we found that 2x-2-200x=0 then wouldnt x be 2/-198 ?

    • 2 years ago
  37. freckles Group Title
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    |dw:1353821740124:dw|

    • 2 years ago
  38. freckles Group Title
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    \[198/2=99\]

    • 2 years ago
  39. freckles Group Title
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    \[2/198=1/99\]

    • 2 years ago
  40. freckles Group Title
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    |dw:1353821859781:dw|

    • 2 years ago
  41. 1ace Group Title
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    omg ok i see !!! i think i got it now !!!

    • 2 years ago
  42. freckles Group Title
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    |dw:1353821881622:dw|

    • 2 years ago
  43. 1ace Group Title
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    thank you !! i get it now !!

    • 2 years ago
  44. freckles Group Title
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    Np. :)

    • 2 years ago
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