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anonymous

  • one year ago

Find the standard form of the equation of the parabola with a focus at (-8, 0) and a directrix at x = 8.

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  1. anonymous
    • one year ago
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    @Hero

  2. Hero
    • one year ago
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    Okay, go ahead and solve it using the formula I gave you previously. This time, the point for the directrix will be (8,y).

  3. anonymous
    • one year ago
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    working it meow

  4. anonymous
    • one year ago
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    I got \[y=x-8\sqrt{2}\]

  5. anonymous
    • one year ago
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    @Hero

  6. Hero
    • one year ago
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    You should show the complete work you did on it. From what I observe, your final answer doesn't seem to have the right form.

  7. anonymous
    • one year ago
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    Im skipping the first step cause It takes so long to write \[\large x^2 + 64 +y^2 = x^2 - 64 +x^2 - y^2\]

  8. anonymous
    • one year ago
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    @Hero \[\large 64 = x^2 - y^2 - 64\] I feel like this is a formula of some sort...

  9. Hero
    • one year ago
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    Skipping steps doesn't sound like such a good idea.

  10. anonymous
    • one year ago
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    I didn't skip any but the first where you plug in the values

  11. anonymous
    • one year ago
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    next i did this: \[\large y^2 = x^2 - 128\]

  12. anonymous
    • one year ago
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    Then took the square root of both sides

  13. Hero
    • one year ago
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    That still doesn't look right.

  14. Hero
    • one year ago
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    You must have made another mistake.

  15. Hero
    • one year ago
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    BTW, \(y^2 + y^2 = 2y^2\)

  16. anonymous
    • one year ago
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    left that out by accident

  17. anonymous
    • one year ago
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    \[\large 64 = -2y^2 +x^2 -64\]

  18. anonymous
    • one year ago
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    Does that look right?

  19. Hero
    • one year ago
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    I wrote out the steps by hand myself. There's a reason why I wanted you to post each step you did just like last time. If you skip steps, you end up confusing yourself. Also, if you never got to the point where you end up using difference of squares, then you may have made yet another mistake.

  20. Hero
    • one year ago
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    I recommend starting over but posting each step this time

  21. anonymous
    • one year ago
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    Alright one sec

  22. anonymous
    • one year ago
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    \[\large (x+8)^2 +y^2 = (x-8)^2 + (x-y)^2\]

  23. anonymous
    • one year ago
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    \[\large then: x^2 +64 +y^2 = x^2 -64 +x^2 - y^2\]

  24. anonymous
    • one year ago
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    Do you see any problems yet? @hero

  25. Hero
    • one year ago
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    Yes, I do actually. Double check your work.

  26. anonymous
    • one year ago
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    \[\large x^2 + 64 +y^2 = x^2 +64 +x^2 +y^2\]

  27. anonymous
    • one year ago
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    Is that correct?

  28. Hero
    • one year ago
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    |dw:1435529904593:dw|

  29. anonymous
    • one year ago
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    oh Sh*t

  30. anonymous
    • one year ago
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    Can't believe I didn't see that earlier

  31. Hero
    • one year ago
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    What didn't you see earlier?

  32. anonymous
    • one year ago
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    in the last parentheses in the formula I put an x instead of a y

  33. anonymous
    • one year ago
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    And when I simplify that down it comes out to \[\large y=\sqrt{-32x}\]

  34. anonymous
    • one year ago
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    unless I did that extremely wrong which I felt like I did

  35. Hero
    • one year ago
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    Yes, because you should have isolated x instead of y.

  36. anonymous
    • one year ago
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    Ok. So what you had was correct

  37. Hero
    • one year ago
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    The parabola is horizontal instead of vertical.

  38. anonymous
    • one year ago
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    Oh yeah XD Derp

  39. anonymous
    • one year ago
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    I have some more if you don't mind :)

  40. anonymous
    • one year ago
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    Different type of question

  41. Hero
    • one year ago
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    Actually, I'm about to log off soon. Others should be available to help.

  42. anonymous
    • one year ago
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    ok thanks @hero

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spraguer (Moderator)
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