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Thread: conic sections

  1. #1
    JROD23
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    conic sections

    1. Write the equation of the parabola given the focus at (-3,-2) and the directrix is the line y=-6. It is helpful to draw a sketch of what is given to help in choosing the correct formula. Be sure you have the correct formula to fill in. leave the equation in standard form.


    2.Given the ellipse equation 9x^2 + 4y^2-18x + 16y=0 , find the coordinates of the center, the vertices, and the foci. Sketch the ellipse on the grid showing each of these points as dots. Show steps for converting to standard form.




    PLEASE INCLUDE WORK STEP BY STEP.
    THANK YOU
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  2. #2
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    Hello, JROD23!

    The ellipse problem is not pretty . . .


    2. Given the ellipse equation: . 9x^2 + 4y^2-18x + 16y\:=\:0
    find the coordinates of the center, the vertices, and the foci.
    Sketch the ellipse.

    To get Standard Form, we must complete-the-square.

    We are given: . 9x^2 - 18x + 4y^2 + 16y \;=\;0

    . . . . . 9(x^2 - 2x\qquad) + 4(y^2 + 4y\qquad) \;=\;0

    . . . . 9(x^2 - 2x + 1 ) + 4(y^2 + 4y + 4 )\; = \;0 + 9 + 16

    . . . . . . . . . . . 9(x-1)^2 + 4(y + 2)^2 \;=\;25

    Divide by 25: . \frac{9(x-1)^2}{25} + \frac{4(y+2)^2}{25} \;=\;1

    And we have: . . \frac{(x-1)^2}{\frac{25}{9}} + \frac{(y+2)^2}{\frac{25}{4}} \;=\;1


    This is a "vertical" ellipse. .Its center is: (1,\,\text{-}2)
    . . The major axis is vertical: . a = \frac{5}{2}
    . . The minor axis is horizontal: . b = \frac{5}{3}

    The vertices (ends of the major axis) are \frac{5}{2} units above and below the center.
    . . Vertices: \left(1,\,\frac{1}{2}\right),\;\left(1,\,\text{-}\frac{9}{2}\right)

    The co-vertices (ends of the minor axis) are \frac{5}{3} units left and right of the center.
    . . Covertices: \left(\text{-}\frac{2}{3},\,\text{-}2\right),\;\left(\frac{8}{3},\,\text{-}2\right)


    The foci are above and below the center. .We must find c.

    We have: . c^2 \:=\:a^2-b^2\quad\Rightarrow\quad c^2 \:=\:\left(\frac{5}{2}\right)^2 - \left(\frac{5}{3}\right)^2 \:=\:\frac{125}{36} \quad\Rightarrow\quad c \,=\,\frac{5\sqrt{5}}{6}

    . . Foci: \left(1,\,\text{-}2+\frac{5\sqrt{5}}{6}\right),\:\left(1,\,\text{-}2-\frac{5\sqrt{5}}{6}\right)

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