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Babynini

  • one year ago

Writing equations for parabolas

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  1. Babynini
    • one year ago
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    Write eq in standard form of the parabola that satisfies f= (4,3) directix x=-1

  2. Babynini
    • one year ago
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    so vertex would be = (1.5,3) ?

  3. Babynini
    • one year ago
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    @zepdrix :)

  4. Babynini
    • one year ago
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    @jim_thompson5910 haha sorry for bugging you all day!

  5. anonymous
    • one year ago
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    |dw:1433483905158:dw|

  6. Babynini
    • one year ago
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    A faster way to do it is realize that we can pick a point on the directrix say d = (-1,3) and the f = (4,3) and then find the middle distance from -1 to 4 which is 1.5 so the v = (1.5,3)

  7. Babynini
    • one year ago
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    I just need to write the equation now. Which i'm not sure how to do D:

  8. zepdrix
    • one year ago
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    mmm vertex looks good :)

  9. Babynini
    • one year ago
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    it's horizontal so: 4p(x-h)=(y-k)^2 4p(x-1.5)=(y-3)^2 ?

  10. Babynini
    • one year ago
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    how do I find p?

  11. Babynini
    • one year ago
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    oh! p is the distance is from vertex to foci?

  12. zepdrix
    • one year ago
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    distance between focus and directrix was 5, ya? p is half of that. the value you used to find the vertex. yes.

  13. Babynini
    • one year ago
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    so p = 2.5 yaya

  14. Babynini
    • one year ago
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    4(2.5)(x-1.5)=(y-3)^2

  15. Babynini
    • one year ago
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    how would I write that more attractively :P

  16. Babynini
    • one year ago
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    10(x-1.5)=(y-3)^2

  17. Babynini
    • one year ago
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    is it ok to write it as y^2=10x-18 ?

  18. Babynini
    • one year ago
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    or should I write it as y= +/- sqroot 10x+-18 ?

  19. anonymous
    • one year ago
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    \[(x+1)^2=(x-4)2+(y-3)^2\] \[x^2+2x+1=x^2-8x+16+(y-3)^2\] \[x^2+2x+1-x^2+8x-16=(y-3)^2\] \[(y-3)^2=10x-15\]

  20. Babynini
    • one year ago
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    @surjithayer we forgot the shift upwards. o.0

  21. zepdrix
    • one year ago
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    shift upwards? what? 0_o

  22. zepdrix
    • one year ago
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    That's what the y-3 is, ya?

  23. Babynini
    • one year ago
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    yeah

  24. Babynini
    • one year ago
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    the focal point is (4,3) so we know that it's not on the x = 0 line

  25. Babynini
    • one year ago
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    but when I put the equation into a graphing thing it puts it on the x line

  26. zepdrix
    • one year ago
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    it does? 0_o https://www.desmos.com/calculator/wyivsbib3n looks ok to me.

  27. Babynini
    • one year ago
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    hrrm I simplified it further and it was coming out wrong. Probably my calculations were wrong. Ok, all good. Thanks :)

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