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Math Help - Transformations of conics?

  1. #1
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    Transformations of conics?

    Write an equation of the translated or rotated graph in general form.

    y=3x^2-2x+5 T(2,-3)

    I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.
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  2. #2
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    Quote Originally Posted by homeylova223 View Post
    Write an equation of the translated or rotated graph in general form.

    y=3x^2-2x+5 T(2,-3)

    I know the graph is a parabola. But I am not sure where I would "plug in" h and k into this equation. To get the translated equation.
    1. You know

    \left|\begin{array}{l}\bar{x} = x+2 \\ \bar{y} =y-3\end{array}\right. ~\implies~ \left|\begin{array}{l}\bar{x}-2 = x \\ \bar{y}+3 =y\end{array}\right.

    2. Plug in the terms for x and y and change the variable:

    y+3 = 3(x-2)^2-2(x-2)+5

    Expand and collect like terms.
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    I am having difficulty with one more question

    y^2+8x=0 rotated on pi/6

    I did this

    x= cos(pi/6)x+sin(pi/6)y
    square root 3 / 2 x+1/2y

    For y I got -1/2X+ square root 3/2y

    Then I plug it in

    (-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)


    But I am having trouble figuring out the equation.
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  4. #4
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    Quote Originally Posted by homeylova223 View Post
    I am having difficulty with one more question

    y^2+8x=0 rotated on pi/6

    I did this

    x= cos(pi/6)x+sin(pi/6)y
    square root 3 / 2 x+1/2y

    For y I got -1/2X+ square root 3/2y

    Then I plug it in

    (-1/2x+ square root 3/2y)^2+8( square root 3 /2 x +1/2y)


    But I am having trouble figuring out the equation.
    1. Do yourself (and us) a favour and start a new thread if you have a new question.

    2. I assume that the center of rotation is the origin(?). If so you know:

    \left|\begin{array}{l}\bar{x}  = x \cdot  \cos\left(\frac{\pi}6\right) - y \cdot \sin\left(\frac{\pi}6\right) \\ \bar{y} = x \cdot  \sin\left(\frac{\pi}6\right) + y \cdot \cos\left(\frac{\pi}6\right) \end{array}\right.

    With \cos\left(\frac{\pi}6\right)=\frac12 \sqrt{3} and \sin\left(\frac{\pi}6\right) = \frac12

    you'll get:

    \left|\begin{array}{l}x = \frac12 ( \bar{x}  \sqrt{3} + \bar{y})  \\ y= \frac12 (\bar{y} \sqrt{3} - \bar{x}  )\end{array}\right.

    3. Plug in these terms into the original equation which yields:

    \frac14 (\bar{y} \sqrt{3} - \bar{x})^2+ 4( \bar{x}  \sqrt{3} + \bar{y}) = 0

    4. Expand the brackets and collect like terms:

    x^2 - 2√3xy + 16√3x + 3y^2 + 16y = 0
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  5. #5
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    I have a question when you wrote the equation why is it square root of 3 instead of square root of 3 over 2? I mean it makes sense in the final answer.

    Also would instead of having 1/2 in the equation would you not have 2 because if you plug it back into the equation with two it makes more sense to me

    As I checked and found out the answer to this problem is

    x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y

    And if you use 2 instead of 1/2 in the equation I would get that answer.
    Last edited by homeylova223; May 17th 2011 at 11:01 AM.
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  6. #6
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    Quote Originally Posted by earboth View Post
    ...
    4. Expand the brackets and collect like terms:

    x^2 - 2√3xy + 16√3x + 3y^2 + 16y = 0
    Quote Originally Posted by homeylova223 View Post
    ...
    As I checked and found out the answer to this problem is

    x^2-2square root 3 xy +3(y)^2+ 16square root 3 x+16y

    And if you use 2 instead of 1/2 in the equation I would get that answer.
    You probably have noticed that I made a mistake typing the equation (I used √3 instead of \sqrt{3}) So Here is the correct equation:

    x^2 - 2\cdot \sqrt{3} \cdot xy + 16\cdot \sqrt{3} \cdot x + 3y^2 + 16y = 0

    Sorry for the confusion.
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