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Math Help - Vertices/Foci of an ellipse

  1. #1
    Junior Member Greymalkin's Avatar
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    Vertices/Foci of an ellipse

    Should be an easy one: Find foci and vertices
    4x2+9y2=1

    answers are
    Vertices= ({{\frac{-1}{2}},0})({{\frac{1}{2}},0})

    Foci= ({\frac{-\sqrt{5}}{6}},0)({\frac{\sqrt{5}}{6}},0)

    I cant seem to be able to turn the equation into proper ellipse form to derive numbers from.
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  2. #2
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    Re: Vertices/Foci of an ellipse

    Hello, Greymalkin!

    \text{Find foci and vertices: }\:4x^2 +9y^2 \:=\:1

    \text{Answers are:}

    . . \text{Vertices: }\left(\pm\tfrac{1}{2},\,0\right)

    . . . \text{Foci: }\left(\pm\tfrac{\sqrt{5}}{6},\,0\right)

    \text{I can't seem to be able to turn the equation into proper ellipse form to derive numbers.}

    The general form is: . \frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1

    Note the coefficient of x^2 should be 1.


    We have: . 4x^2 + 9y^2 \:=\:1

    Multiply the first term by \frac{\frac{1}{4}}{\frac{1}{4}}, the second by \frac{\frac{1}{9}}{\frac{1}{9}}

    . . \frac{\frac{1}{4}}{\frac{1}{4}}\left(\frac{4x^2}{1  }\right) + \frac{\frac{1}{9}}{\frac{1}{9}}\left(\frac{9y^2}{1  }\right) \:=\:1 \quad\Rightarrow\quad \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} \:=\:1


    Hence: . \begin{Bmatrix}a^2 \:=\:\frac{1}{4} & \Rightarrow & a \:=\:\pm\frac{1}{2} \\ \\[-3mm] b^2 \:=\:\frac{1}{9} & \Rightarrow & b \:=\:\pm\frac{1}{3} \end{Bmatrix}

    And: . c^2 \:=\:a^2-b^2 \:=\:\tfrac{1}{4} - \tfrac{1}{9} \:=\:\tfrac{5}{36} \quad\Rightarrow\quad c \:=\:\pm\tfrac{\sqrt{5}}{6}
    Thanks from Greymalkin
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