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Thread: Graph the ellipse and graph the hyperbola

  1. #1
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    Unhappy Graph the ellipse and graph the hyperbola

    Please help me solve the following ellipse problem:

    Graph the ellipse. 16x^2+4y^2=48

    what is the center?
    Where are the vertices?
    Where are the co-vertices?
    Where are the foci?




    Graph the hyperbola. 16y^2-4x^2=64
    What is the center?
    What are the vertices?
    Where are the co-vertices?
    Where are the foci?
    What are the asymptotes?
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  2. #2
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    Quote Originally Posted by smartcp View Post
    Graph the ellipse. 16x^2+4y^2=48

    what is the center?
    Where are the vertices?
    Where are the co-vertices?
    Where are the foci?
    Most folks resort to a Standard Form, usually this one

    $\displaystyle \frac{x^{2}}{3} + \frac{y^{2}}{12} = 1$

    Using standard notation:

    This makes $\displaystyle a = \sqrt{3}$, and $\displaystyle b = 2\sqrt{3}$

    Find 'c' and you nearly are done.

    Note: If you REALLY cannot find the center of this one, you MUST go back and reread a section or two. You missed something very important.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by smartcp View Post
    Please help me solve the following ellipse problem:

    Graph the ellipse. 16x^2+4y^2=48

    what is the center?
    Where are the vertices?
    Where are the co-vertices?
    Where are the foci?




    Graph the hyperbola. 16y^2-4x^2=64
    What is the center?
    What are the vertices?
    Where are the co-vertices?
    Where are the foci?
    What are the asymptotes?
    Graph of Ellipse:


    I'll let you figure out the rest.

    Graph of Hyperbola:



    Try to figure out the rest.
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  4. #4
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    Quote Originally Posted by smartcp View Post
    Please help me solve the following ellipse problem:

    Graph the ellipse. 16x^2+4y^2=48

    what is the center?
    Where are the vertices?
    Where are the co-vertices?
    Where are the foci?




    Graph the hyperbola. 16y^2-4x^2=64
    What is the center?
    What are the vertices?
    Where are the co-vertices?
    Where are the foci?
    What are the asymptotes?
    1.)
    $\displaystyle 16x^2+4y^2=48$
    $\displaystyle \frac{x^2}{3} + \frac{y^2}{12} = 1$


    2.)
    $\displaystyle 16y^2-4x^2=64$
    $\displaystyle \frac{y^2}{4} - \frac{x^2}{16}=1$


    Standard form: $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
    Vertices: $\displaystyle (\pm a, 0)$
    Co-vertices: $\displaystyle (0, \pm b)$
    Eccentricity: $\displaystyle e<1, \ \ b^2 = a^2(1-e^2)$
    Foci: $\displaystyle (\pm ae, 0)$
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