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Math Help - Asymptotes for hyperbolas

  1. #1
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    Asymptotes for hyperbolas

    Find the asymptotes for the hyperbola  <br />
\frac{2x^2}{3} - \frac{y^2}{2}=1<br />

    I got the following:
     <br />
y=\frac{\sqrt{6}}{3}<br />
and y=\frac{-\sqrt{6}}{3}

    However the answer is y=\pm \frac{4}{\sqrt{3}}

    Where did I go wrong? Please, any help would be appreciated!
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  2. #2
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    Quote Originally Posted by Joker37 View Post
    Find the asymptotes for the hyperbola  <br />
\frac{2x^2}{3} - \frac{y^2}{2}=1<br />

    I got the following:
     <br />
y=\frac{\sqrt{6}}{3}<br />
and y=\frac{-\sqrt{6}}{3}

    However the answer is y=\pm \frac{4}{\sqrt{3}}

    Where did I go wrong? Please, any help would be appreciated!
    \displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}

    \Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}

    and as x\rightarrow\infty, this is \displaystyle\ y=\pm\frac{2x}{\sqrt{3}}
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  3. #3
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    From Archie's post to know where the function is undefined solve for \displaystyle 4x^2-6<0 \implies -\frac{\sqrt{3}}{\sqrt{2}}<x<\frac{\sqrt{3}}{\sqrt{  2}}
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  4. #4
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    Do you understand that it is impossible to say "where you went wrong" if you don't show what you did?

    In any case, a hyperbola having the coordinate axes as axes of symmetry (any hyperbola of the form \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1 or \frac{y^2}{b^2}- \frac{x^2}{a^2}= 1) has slant asymptotes, not vertical and horizontal asymptotes.

    No, the anwer is not y= \pm\frac{4}{\sqrt{3}}. Neither y= \pm\frac{\sqrt{6}}{3} nor y= \pm\frac{4}{\sqrt{3}} is correct.

    Here is one way to think about it. The graph of the hyperbola gets closer and closer to the asymptotes as x and y get larger and larger. For very, very large x and y, both \frac{2x^2}{3} and \frac{y^2}{2} are very, very large so that "1" can be ignored compared to them. That is, for very, very large x and y, \frac{2x^2}{3}- \frac{y^2}{2}= 0, approximately. The left side is a "difference of squares" and factors as \left(\frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}\right)\left(\frac{x\sqrt{2}}{\s  qrt{3}}- \frac{y}{\sqrt{2}}\right)= 0.

    So we must have \frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0 and \frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0, the equation of the two straight line asymptotes. Solving for y, we have y= \pm\frac{2}{\sqrt{3}}x as Archie Meade says.
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}

    \Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}

    and as x\rightarrow\infty, this is \displaystyle\ y=\pm\frac{2x}{\sqrt{3}}
    Technically, the generalised binomial theorem should be used to find this limit.
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  6. #6
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    Hello, Joker37!

    \text{Find the asymptotes for the hyperbola: }\:\dfrac{2x^2}{3} - \dfrac{y^2}{2}\:=\:1

    We're expected to be familiar with this fact:

    Given the hyperbola: . \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} \:=\:1, the asymptotes are: . y \;=\;\pm\dfrac{b}{a}x



    The equation is: . \displaystyle \frac{x^2}{\frac{3}{2}} - \frac{y}{2} \;=\;1 \quad\Rightarrow\quad \frac{x^2}{\left(\sqrt{\frac{3}{2}}\right)^2} - \frac{y^2}{(\sqrt{2})^2} \;=\;1

    Hence: . a = \sqrt{\frac{3}{2}},\;b = \sqrt{2}


    Therefore: . y \;=\;\pm\dfrac{\sqrt{2}}{\sqrt{\frac{3}{2}}}x \quad\Rightarrow\quad y \;=\;\pm\dfrac{2}{\sqrt{3}}x

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