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Thread: 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
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    Hello, Justin!

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

    1) Find the center and vertices of the hyperbola.

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

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

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

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

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



    2) Write the equation in general form

    Your algebra is faulty . . .

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

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

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

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

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

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