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Math Help - confusing math on conics

  1. #1
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    Wink confusing math on conics

    Given (y-2)^2 - X^2/4=1, do the following

    a) write the equation in general form:

    my answer:
    (y-2)^2 - X^2/4=4
    4(y-2)^2-1(x)^2=4
    4(y^2-4y+4) -1(x^2+x)=4
    4y^2-16y+16-(x^2+x)= 4
    4y^2-16y+16-X^2+x=4
    -x^2+4y^2+x-16y-16=4
    -x^2+4y^2+x-16y-20=0

    general form:
    -x^2+4y^2+x-16y-20=0

    How do I locate the vertices and center of this hyperbola??
    thanks for your help!!!
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Justin!

    Given: . (y-2)^2 - \frac{x^2}{4}\:=\:1

    1) Find the center and vertices of the hyperbola.

    I must ask: Do you know anything about the hyperbola?

    We have: . \frac{(y-2)^2}{1} - \frac{x^2}{4}\:=\:1

    The center is: (0,2). . a = 2,\:b = 1

    The hyperbola is "vertical": \begin{array}{ccc}\lor \\ \land\end{array}

    The vertices are one unit above and one unit below the center.
    . . V\!: \;(0,3),\;(0,1)



    2) Write the equation in general form

    Your algebra is faulty . . .

    We have: . \frac{(y-2)^2}{1} - \frac{x^2}{4}\:=\:1

    Multiply by 4\!:\;\;4(y-2)^2 - x^2\;=\;4

    . . 4(y^2 - 4y + 4) - x^2\;=\;4

    . . 4y^2 - 16y + 16 - x^2\:=\:4

    . . -x^2 + 4y^2 - 16y + 12\;=\;0\;\;\text{ or }\;\;x^2 - 4y^2 + 16y - 12\;=\;0

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