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1ace

  • 3 years ago

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

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  1. freckles
    • 3 years ago
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    You know the distance formula?

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

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

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

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

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

  7. freckles
    • 3 years ago
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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)

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

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

  10. 1ace
    • 3 years ago
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    yep i did that ! :o

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

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

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

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

  15. 1ace
    • 3 years ago
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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

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

  17. freckles
    • 3 years ago
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    We know that (100)'=0

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

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

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

  21. 1ace
    • 3 years ago
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    okay!

  22. freckles
    • 3 years ago
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    Right!: )

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

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

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

  26. freckles
    • 3 years ago
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    No no.

  27. 1ace
    • 3 years ago
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    why not ?

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

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

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

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

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

  33. freckles
    • 3 years ago
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    Say yes. lol

  34. 1ace
    • 3 years ago
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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

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

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

  37. freckles
    • 3 years ago
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    |dw:1353821740124:dw|

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

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

  40. freckles
    • 3 years ago
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    |dw:1353821859781:dw|

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

  42. freckles
    • 3 years ago
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    |dw:1353821881622:dw|

  43. 1ace
    • 3 years ago
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    thank you !! i get it now !!

  44. freckles
    • 3 years ago
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    Np. :)

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