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    I'm confused especially where it says sketch the curveS - is it expecting me to sketch y=x^2-2x+1 and y^2+2y?

    Can this not just be solved by writing it in the Cartesian form for a circle, and the point is where the circle touches the x axis?

    ------------------------



    Here's what I did.

    The shortest distance will be the normal to the parabola.

    y'(x)=-x

    So grad of normal is:

    \frac{1}{x}

    And it passes through the origin so the line is:

    y-0=\frac{1}{x}(x-0)

    y=1

    (confused here, shouldn't it be y=ax?)

    But anyway, subbing that value for y into our y(x)


    x^2=2

    x\pm\sqrt{2}

    Which gives y values of both 1.

    Meaning the line segment is..

    l=\sqrt{2+1}

    l=\sqrt{3}

    Is this right? Can anyone confirm?
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    (Original post by SamKeene)

    I'm confused especially where it says sketch the curveS - is it expecting me to sketch y=x^2-2x+1 and y^2+2y?

    Can this not just be solved by writing it in the Cartesian form for a circle, and the point is where the circle touches the x axis?

    ------------------------



    Here's what I did.

    The shortest distance will be the normal to the parabola.

    y'(x)=-x

    So grad of normal is:

    \frac{1}{x}

    And it passes through the origin so the line is:

    y-0=\frac{1}{x}(x-0)

    y=1

    (confused here, shouldn't it be y=ax?)

    But anyway, subbing that value for y into our y(x)


    x^2=2

    x\pm\sqrt{2}

    Which gives y values of both 1.

    Meaning the line segment is..

    l=\sqrt{2+1}

    l=\sqrt{3}

    Is this right? Can anyone confirm?
    although I did it differently i get exactly the same
 
 
 
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