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Math Help - focus of the parabola that has vertex (−3, 2),passes through the point (−10, 1).

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    focus of the parabola that has vertex (−3, 2),passes through the point (−10, 1).

    Find the focus of the parabola that has vertex (−3, 2), opens horizontally, and passes through the point (−10, 1).

    (−83/28, 2)(−87/28, 2) (−85/28, 2) (−4, 2)None of the above
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  2. #2
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    Help me out please
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  3. #3
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    any hints?
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    For other members, please note that this is apparently an exam question (see here). I urge you not to encourage academic dishonesty by offering direct assistance on this problem.
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    I almost didn't view this thread, because your various replies made it look as though it had been answered already!

    Quote Originally Posted by skylitejdm View Post
    Find the focus of the parabola that has vertex (−3, 2), opens horizontally, and passes through the point (−10, 1).
    To learn the basic process for finding the equation from some provided information, try here.

    You know that the equation is something along the lines of 4p(x - h) = (y - k)^2, because the parabola opens horizontally instead of vertically.

    You are given that (h, k) = (-3, 2). You are also given that the point (-10, 1) lies on the curve. Then:

    . . . . .4p(x + 3) = (y - 2)^2

    Plugging in the given point:

    . . . . .4p(-10 + 3) = (1 - 2)^2

    . . . . .4p(-7) = (-1)^2

    . . . . .-28p = 1

    . . . . .p = -1/28

    Since p is negative, then the parabola opens to the left, with the vertex being on the right. To which side of the vertex, then, is the focus? How far is the focus from the vertex?

    Then what are the coordinates of the focus?

    If you get stuck, please reply showing all of your work and reasoning so far.

    Thank you!
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    Quote Originally Posted by Reckoner View Post
    For other members, please note that this is apparently an exam question (see here). I urge you not to encourage academic dishonesty by offering direct assistance on this problem.
    thank you for not letting me learn an equation like this and for not giving me the opportunity to give donation for people that can help me.... WOW is it a crime to get help from this forum? it that the whole purpose why they even made this forum!
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    Um... You might want to review those terms and conditions, etc. They're pretty explicit about wanting to aid learning, not facilitate cheating.
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    Quote Originally Posted by skylitejdm View Post
    thank you for not letting me learn an equation like this and for not giving me the opportunity to give donation for people that can help me.... WOW is it a crime to get help from this forum? it that the whole purpose why they even made this forum!
    Read the Forum Rules:

    10) Do not beg for answers. If your question is unanswered do not make a post begging for someone to reply back, it is totally useless. Other members can see the question.

    20) MHF does not endorse academic dishonesty in any way. Do not attempt to get help with a take-home test, exam or anything of that nature. Certain teachers and professors allow help with general homework and some do not. Make sure you find out if you are allowed to receive help. Those caught cheating will not only be banned, but we will also do everything we can to report you to your academic institution. We have done so in the past and will do so again.
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    thanks for the advise... i turn the test in already... just trying to learn it now
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