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Math Help - need help with hyperbola and ellipse

  1. #1
    lvc
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    need help with hyperbola and ellipse

    i have the eqautions

    5x^2 + 9y^2=45

    16x^2-4y^2+64=0


    how do i make these into the hyperbola and ellipse form
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  2. #2
    Master Of Puppets
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    5x^2 + 9y^2=45

    \frac{5x^2}{5} + \frac{9y^2}{5}=\frac{45}{5}

    x^2 + \frac{9y^2}{5}=9

    \frac{x^2}{9} + \frac{9y^2}{9\times 5}=\frac{45}{9\times 5}

    \frac{x^2}{9} + \frac{y^2}{ 5}=1

    \frac{x^2}{\sqrt{9}^2} + \frac{y^2}{\sqrt{ 5}^2}=1
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  3. #3
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    Hello, lvc!

    16x^2-4y^2+64\:=\:0

    We have: . 4y^2 - 16x^2 \:=\:64

    Divide by 64: . \displaystyle \frac{4y^2}{64} - \frac{16x^2}{64} \;=\;\frac{64}{64} \quad\Rightarrow\quad \frac{y^2}{16} - \frac{x^2}{4} \;=\;1


    (In the first one, just divide by 45.)

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  4. #4
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    Quote Originally Posted by lvc View Post
    i have the eqautions

    5x^2 + 9y^2=45

    16x^2-4y^2+64=0


    how do i make these into the hyperbola and ellipse form
    They are already in "hyerbola and ellipse form". What you mean is into "standard form".

    Standard form for an ellipse is \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1
    and you have 5x^2+ 9x^2= 45. Divide both sides by 45 to get "1" on the right.

    Standard form for an ellipse is either \frac{x^2}{a^2}- \frac{y^2}{b^2}= 1 or \frac{y^2}{b^2}- \frac{x^2}{a^2}= 1. You have 16x^2- 4y^2+ 64= 0 which is the same as 16x^2- 4y^2= -64. (We cannot divide "0" by anything to get "1" so I subtracted 64 from both sides.) Divide both sides by -64 to get "1" on the right.
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