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    4. The parabola y2 = 8x is transformed by a vertical stretch, scale factor 2, and then a horizontal translation of -1 unit. Write down the equation of the resulting parabola in the form y = f(x).
    (this is not an exam question)

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    Answer is
    y = 4 √2(x+1)

    What did I screw up this time?
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    i would make y the subject first ?

    y = ± √ ( 8x )

    then stretch

    y = ± 2√ ( 8x )

    then shift to the left

    y = ± 2√ ( 8{x + 1} )
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    (Original post by ckfeister)
    4. The parabola y2 = 8x is transformed by a vertical stretch, scale factor 2, and then a horizontal translation of -1 unit. Write down the equation of the resulting parabola in the form y = f(x).
    (this is not an exam question)

    Answer is
    y = 4 √2(x+1)

    What did I screw up this time?
    Should be \displaystyle (\frac{y}{2})^2=8(x+1) instead. Your working suggests it's a vertical stretch by a factor of 1/2.
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    (Original post by RDKGames)
    Should be \displaystyle (\frac{y}{2})^2=8(x+1) instead. Your working suggests it's a vertical stretch by a factor of 1/2.


    1.a) Sketch the ellipse   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where a = 3 and b = 2

    b) Write down (i) the length of the major axis

    (ii) the length of the minor axis

    c) Draw the line y = x on your sketch.

    d) Find the co-ordinates of the points of intersection of the line and the ellipse.

    e) Reflect the ellipse in the line y = x and write down the equation of the resulting ellipse.



    I'm on d)
    Somehow the x values are x = 2,3... how?
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    (Original post by ckfeister)
    1.a) Sketch the ellipse   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where a = 3 and b = 2

    b) Write down (i) the length of the major axis

    (ii) the length of the minor axis

    c) Draw the line y = x on your sketch.

    d) Find the co-ordinates of the points of intersection of the line and the ellipse.

    e) Reflect the ellipse in the line y = x and write down the equation of the resulting ellipse.



    I'm on d)
    Somehow the x values are x = 2,3... how?
    Because \sqrt{4x^2+9y^2}\not= 2x+3y. Take x=y=1 as a counterexample.
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    (Original post by RDKGames)
    Because \sqrt{4x^2+9y^2}\not= 2x+3y. Take x=y=1 as a counterexample.
     2x^2 + 3y^2 ?
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    (Original post by RDKGames)
    Because \sqrt{4x^2+9y^2}\not= 2x+3y. Take x=y=1 as a counterexample.
    Only different I get is its x = +- square root 5/6
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    (Original post by RDKGames)
    Because \sqrt{4x^2+9y^2}\not= 2x+3y. Take x=y=1 as a counterexample.
    Wait, the answer is x = +-1.6666, not 2,3 but I still get x = +-5/6
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    (Original post by ckfeister)
     2x^2 + 3y^2 ?
    No, that is not how you square root addition of two different variables.

    Since y=x you just have 4x^2+9x^2=36 so just solve for x.


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