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tw101

  • one year ago

Write the equation of the parabola with a focus (1,2) and directrix y = 1.

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  1. tw101
    • one year ago
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    Im not sure how to start.

  2. tw101
    • one year ago
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    1?

  3. tw101
    • one year ago
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    okay the formula you put up above?

  4. tw101
    • one year ago
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    oh okay, onne second and then ill tell you what I got:)

  5. jdoe0001
    • one year ago
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    one may note that 1,2 is the focus point, not the vertex point

  6. tw101
    • one year ago
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    oh wait what does that mean then?

  7. jdoe0001
    • one year ago
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    well... she maybe crazy, but still beautiful, so tis ok =)

  8. tw101
    • one year ago
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    haha.

  9. tw101
    • one year ago
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    im seriously confused though...

  10. jdoe0001
    • one year ago
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    |dw:1439942544203:dw|

  11. tw101
    • one year ago
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    Im not sure... Im sorry, this is new to me.

  12. jdoe0001
    • one year ago
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    hmmm well.. have you covered parabolas yet?

  13. tw101
    • one year ago
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    Yes, briefly.

  14. jdoe0001
    • one year ago
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    well.. it turns out that the focus point is "p" distance from the vertex and the directrix is "p" distance also from the vertex so.. the vertex is really half-way between both, since the focus and directrix are the same distance from the vertex keeping in mind that, the parabola opens up towards the focus point

  15. tw101
    • one year ago
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    Okay, well crazyandbeautiful helped me out and said that (1,3/2) was the vertex. haha.

  16. jdoe0001
    • one year ago
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    hmmm how did she get 1, 3/2 though?

  17. tw101
    • one year ago
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    haha no you dont need to its okay! @Crazyandbeautiful

  18. tw101
    • one year ago
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    okay so its always halfway between the focus point and directrix?

  19. jdoe0001
    • one year ago
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    yeap, they're always "p" distance from the vertex

  20. tw101
    • one year ago
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    Okay, i get that now:)

  21. tw101
    • one year ago
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    Where do I go from there?

  22. jdoe0001
    • one year ago
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    so... get the vertex, get "p" and plug and chug \(\large{ \begin{array}{llll} (y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\ (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}}\)

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