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Math Help - Two central conic questions.

  1. #1
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    Two central conic questions.

    I have two questions for which I have partial working for, but I am unsure of the working for them.

    Q)1:
    Find (a) the directrix, (b) the focus and (c) the roots of the parabola y=x^2-5x+4

    Answer(s):
    a)
    y= -\frac{5}{2}

    b)
    (2.5,-2)

    c)
    y=x^2-5x+4\rightarrow y=(x-4)(x-1)
    Therefore x=4 and x=1.

    I'm very certain of my answer for c), however I lost my working for a) and b) so I'm fairly certain that I am incorrect for those answers.

    Q)2
    Graph  \frac{x^2}{16}-\frac{y^2}{25}=1
    Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asympotes.

    A) I admit to being lost when it comes to this question, I know how to work these kind of questions under other situations where the x and y are reversed. E.g.: \frac{y^2}{25}-\frac{x^2}{16}. I am very lost as to what to do with this one though.
    When I try to determine 'c' I come out with \sqrt{41}
    I don't really see how that can be correct.

    Thank you.
    Last edited by quikwerk; January 27th 2011 at 09:18 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Completing squares:

    y=\left(x-5/2\right)^2-25/4+4

    or

    y-(-9/4)=4(1/4)(x-5/2)^2

    we have the parabola in the canonical form:

    (y-\beta)=4f(x-\alpha)^2

    Then, the directrix is:

    y=\beta-f=-9/4-1/4=-5/2

    etc.


    Fernando Revilla
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  3. #3
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    I'm not at all expireienced with canonical form of parabolas...so I dont really understand what you are saying...
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by quikwerk View Post
    I'm not at all expireienced with canonical form of parabolas...so I dont really understand what you are saying...

    Well, that is a standard way. For helping you I should know what can I use.


    Fernando Revilla


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  5. #5
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    I can use completing the square and other high school algebra methods.
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by quikwerk View Post
    I'm not at all expireienced with canonical form of parabolas...so I dont really understand what you are saying...
    Hi quikwerk,

    The standard (or vertex) form of your parabola is y=a(x-h)^2+k

    where (h, k) is your vertex.

    The directrex is y=k-\frac{1}{4a}

    To put your equation into vertex form, you complete the square like FernandoRevilla showed you.

    y=x^2-5x+4

    y=(x^2-5x+\frac{25}{4})+4-\frac{25}{4}

    y=(x-\frac{5}{2})^2-\frac{9}{4}

    Your directrix is y=-\frac{9}{4}-\frac{1}{4(1)}

    y=-\frac{5}{2}


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  7. #7
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    Many thanks, both of you!
    Masters, I finally understand how to work this question.

    Can you help me with the second question?

    I have determined the x-intercepts to be 4 and -4 and I have figured out the asymptotes as y=\pm\frac{5x}{4}
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  8. #8
    A riddle wrapped in an enigma
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    The focus in Q1 can be found using this little ditty:

    F\left(h\;,\;k+\frac{1}{4a}\right)

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  9. #9
    A riddle wrapped in an enigma
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    Quote Originally Posted by quikwerk View Post
    Many thanks, both of you!
    Masters, I finally understand how to work this question.

    Can you help me with the second question?

    I have determined the x-intercepts to be 4 and -4 and I have figured out the asymptotes as y=\pm\frac{5x}{4}
    Post Q2 as a separate post and someone will get to it. I promise. I'm going to be tied up for about an hour.
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  10. #10
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    Thanks again Masters!
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