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Math Help - Centeral conic question.

  1. #1
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    Centeral conic question.

    I'm not too sure of how to work this one out, I'm really having trouble determining the extent of the graph. I have figured out that the x intercept is \pm4 and that the asymptotes are y=\frac{5x}{4}and -\frac{}{5x}{4}
    Graph
    Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asymptotes.

    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.: . I am very lost as to what to do with this one though.
    When I try to determine 'c' I come out with
    I don't really see how that can be correct.

    Thank you.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by quikwerk View Post
    I'm not too sure of how to work this one out, I'm really having trouble determining the extent of the graph. I have figured out that the x intercept is \pm4 and that the asymptotes are y=\frac{5x}{4}and -\frac{}{5x}{4}
    Graph
    Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asymptotes.

    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.: . I am very lost as to what to do with this one though.
    When I try to determine 'c' I come out with
    I don't really see how that can be correct.

    Thank you.
    Hi quikwerk,

    You are correct. c=\sqrt{41}

    The standard form for this hyperbola is \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

    The equations of the asymptotes are y-k=\pm \frac{b}{a}(x-h)

    y-0=\pm \frac{5}{4}(x-0)

    y=\pm \frac{5}{4}x
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  3. #3
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    Thanks- still not sure how to work out the extent of the graph : (

    Am i correct in thinking that the extent of the graph is given by the equations y=\pm\frac{5}{4}\sqrt{x^2-16}and x=\pm\frac{4}{5}\sqrt{y^2-25}?
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  4. #4
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    Quote Originally Posted by quikwerk View Post
    Thanks- still not sure how to work out the extent of the graph : (

    Am i correct in thinking that the extent of the graph is given by the equations y=\pm\frac{5}{4}\sqrt{x^2-16}and x=\pm\frac{4}{5}\sqrt{y^2-25}?
    My guess would be the extents of the graph would be the Domain and Range.
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  5. #5
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    Quote Originally Posted by masters View Post
    My guess would be the extents of the graph would be the Domain and Range.

    Does that mean that \mid x \mid\geq 4 and \mid y \mid\geq 5?
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  6. #6
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    Quote Originally Posted by quikwerk View Post
    Does that mean that \mid x \mid\geq 4 and \mid y \mid\geq 5?
    The Domain would be r\:x \ge 4\}" alt="\{x\;|\;x \le -4 \r\:x \ge 4\}" />

    The Range would be \{y\;|\; y \in \Re\}
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  7. #7
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    Two points: To find the "asymptotes", think about what happens for x and y really, really big! For \frac{x^2}{16} and \frac{y^2}{25} are far larger than the "1" on the right side- if we ignore, it we see that the curve must be close to \frac{x^2}{16}- \frac{y^2}{25}= \left(\frac{x}{4}- \frac{y}{5}\right)\left(\frac{x}{4}+ \frac{y}{5}\right)= 0 which means that either \frac{x}{4}- \frac{y}{5}= 0 so that 4y= 5x or \frac{x}{4}+ \frac{y}{5}= 0 so that 4y= -5x. Those are the asymptotes.

    For the "extent" try solving for x and y. \frac{y^2}{25}= \frac{x^2}{16}- 1 so that y= \pm 5\sqrt{\frac{x^2}{16}-1}. Since the argument of the square root must be positive, we must have \frac{x^2}{16}- 1\ge 0 or x^2\ge 16 so that x\le -4 or x\ge 4.
    x= \pm 4\sqrt{\frac{y^2}{25}+ 1}. Since the argument is always positive, there is no restriction on y.
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  8. #8
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    Thank you guys! I'll try to work it out with your advice and report back later.
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