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Math Help - Conics & Calculus: Conics, Parametric, and Polar Coordinates

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    Conics & Calculus: Conics, Parametric, and Polar Coordinates

    Hi, I pealse need an explanation for the following problems. I have no clue..

    a)Find the vertex, focus, and directrix of the parabola of the follwing exercices and sketh the graph of the parabola.

    1) x^2 +8y = 0

    2) (x-1)^2 + 8 (y+2) = 0

    3) y^2 + 6y + 8x + 25 = 0


    b) Find an equation of parabola of the following exercices

    4) vertex: (-1, 2)
    focus: (-1,0)

    5) focus: (2, 2)
    directrix: x= -2

    6) When axis is parallel to y-axis; and when graph passes through (0, 3) , (3, 4), and (4, 11).
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by googoogaga View Post
    Hi, I pealse need an explanation for the following problems. I have no clue..

    a)Find the vertex, focus, and directrix of the parabola of the follwing exercices and sketh the graph of the parabola.

    1) x^2 +8y = 0

    2) (x-1)^2 + 8 (y+2) = 0

    3) y^2 + 6y + 8x + 25 = 0


    b) Find an equation of parabola of the following exercices

    4) vertex: (-1, 2)
    focus: (-1,0)

    5) focus: (2, 2)
    directrix: x= -2
    this post may help. it will at least familiarize you with the rules you need to know
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by googoogaga View Post

    6) When axis is parallel to y-axis; and when graph passes through (0, 3) , (3, 4), and (4, 11).
    i'm sure there's an easier way, but i'd probably approach this with simultaneous equations.

    Let the desired parabola be of the form y = ax^2 + bx + c

    Since it passes through (0,3), when x = 0, y = 3. so we have:
    3 = a(0)^2 + b(0) + c = c................(1)

    \Rightarrow c = 3, we can use this in the other equations right off the bat, nice!

    Since it passes through (3,4), when x = 3, y = 4. so we have:
    4 = a(3)^2 + b(3) + 3 , since c = 3

    \Rightarrow 9a + 3b = 1 .....................(2)


    Since it passes through (4,11), when x = 4, y = 11. so we have:
    11 = a(4)^2 + b(4) + 3

    \Rightarrow 16a + 4b = 8 ....................(3)


    So now we have the system:

    9a + 3b = 1 .................(2)
    4a + b = 2 ...................(3)

    Now solve that system and plug in the values for a,b, and c into the form of the quadratic to get your answer
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