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anonymous

  • one year ago

Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?

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  1. AlexandervonHumboldt2
    • one year ago
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    Focus: (2, 6) Any point, (x0 , y0) on the parabola satisfies the definition of parabola, so there are two distances to calculate: Distance between the point on the parabola to the focus Distance between the point on the parabola to the directrix \[\sqrt{(x_0-a)^2+(y_0-b)^2} \] \[\sqrt{(x_0-2)^2+(y_0-0)^2}\] y=-2 dirextix Distance between point ( x0 , y0) and the line y=-2 |y_0+2| Equate the two expressions. \[\sqrt{(x_0-2)^2+(y_0-0)^2}=\left| y_0+2 \right|\] now solve this:

  2. AlexandervonHumboldt2
    • one year ago
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    square both sides: \[(x_0-2)^2+y_0^2=(y_0+2)^2\]

  3. AlexandervonHumboldt2
    • one year ago
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    simplify: \[x^2-2x+4+y^2=y^2+2y+4\]

  4. AlexandervonHumboldt2
    • one year ago
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    simplify: \[x^2-2x+4-4+y^2-y^2=2y\]

  5. AlexandervonHumboldt2
    • one year ago
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    \[x^2-2x=2y\]

  6. AlexandervonHumboldt2
    • one year ago
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    \[y=\frac{ x^2-2x }{ 2 }\]

  7. AlexandervonHumboldt2
    • one year ago
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    i might have done a mistake though haven't made such question for long time

  8. anonymous
    • one year ago
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    thats not any of my choices :/

  9. AlexandervonHumboldt2
    • one year ago
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    what are your choises?

  10. anonymous
    • one year ago
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    \[f(x)= -\frac{ 1 }{ 8 } (x-2)^{2}-2\]

  11. anonymous
    • one year ago
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    \[f(x)=\frac{ 1 }{ 16 } (x-2)^{2}-2\]

  12. anonymous
    • one year ago
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    and then the same as the first except positive 1/8 and same as the second but negative 1/16

  13. AlexandervonHumboldt2
    • one year ago
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    OHHHHHHHHHHHHH i counted with focus (2, 0) not (2, 6) wait lemme redo it

  14. anonymous
    • one year ago
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    okey

  15. AlexandervonHumboldt2
    • one year ago
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    \[\sqrt{(x_0-2)^2+(y_0-6)^2}=\left| y_0+2 \right|\]

  16. AlexandervonHumboldt2
    • one year ago
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    \[(x_0-2)^2+(y_0-6)^2=(y_0+2)^2\]

  17. AlexandervonHumboldt2
    • one year ago
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    \[(x-2)^2+y^2-12y+36=y^2+2y+4\]

  18. AlexandervonHumboldt2
    • one year ago
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    \[(x-2)^2+32=14y\]

  19. AlexandervonHumboldt2
    • one year ago
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    woops mistake again

  20. AlexandervonHumboldt2
    • one year ago
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    (x-2)^2+y^2-12y+36=y^2+4y+4

  21. AlexandervonHumboldt2
    • one year ago
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    (x-2)^2+32=16y y=1/16*(x-2)^2+2 i the answer if i again didn't made a mistake somewhere

  22. anonymous
    • one year ago
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    thanks! ill let you know in 1 sec

  23. anonymous
    • one year ago
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    yep! it was right!! i got 100/100 on my quiz thanks!

  24. AlexandervonHumboldt2
    • one year ago
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    np sorry for taking so much time

  25. AlexandervonHumboldt2
    • one year ago
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    np sorry for taking so much time

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