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Thread: Give equation in standard form of a parabola

  1. #1
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    Give equation in standard form of a parabola

    Give an equation in standard form of a parabola with vertex (-2,1) and drectrix =-1
    Show work.
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    Re: Give equation in standard form of a parabola

    We don't do homework; show how far you got and we'll help...
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    Re: Give equation in standard form of a parabola

    Quote Originally Posted by Nfalconer View Post
    Give an equation in standard form of a parabola with vertex (-2,1) and drectrix =-1
    Show work.
    First note that the directrix is vertical which means that the axis of symmetry is horizontal.
    The vertex is to the left of the directrix, so the parabola opens to the left.
    The vertex is equally distance from the focus & the directrix on thee axis of symmetry.

    Thus what is the Focus?

    Now apply the definition: A parabola is the locus of points equally distance from its focus and its directrix.
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    Re: Give equation in standard form of a parabola

    I got a focus of -3,1 and equation in standard form as x=-4 (y-1)2nd -2. Don't think it is correct
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    Re: Give equation in standard form of a parabola

    Standard form x=-4(y-1) squared -2. Not sure if it is correct
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    Re: Give equation in standard form of a parabola

    Quote Originally Posted by Nfalconer View Post
    Standard form x=-4(y-1) squared -2. Not sure if it is correct
    Well I get $\large{(y-1)^2=-4(x-2)}$
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    Re: Give equation in standard form of a parabola

    I get something slightly different (from the result given by the OP). For an arbitrary point $\displaystyle (x,y)$ on the parabola, we set the perpendicular distance to the directrix equal to the distance to the focus and square:

    $\displaystyle (x+1)^2=(x+3)^2+(y-1)^2$

    $\displaystyle x^2+2x+1=x^2+6x+9+(y-1)^2$

    $\displaystyle -4x=(y-1)^2+8$

    $\displaystyle x=-\frac{1}{4}(y-1)^2-2$

    Give equation in standard form of a parabola-mhf_0002.png
    Last edited by MarkFL; Jul 7th 2018 at 08:02 PM.
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    Re: Give equation in standard form of a parabola

    Quote Originally Posted by Plato View Post
    Well I get $\large{(y-1)^2=-4(x-2)}$
    Quote Originally Posted by MarkFL View Post
    I get something slightly different (from the result given by the OP). For an arbitrary point $\displaystyle (x,y)$ on the parabola, we set the perpendicular distance to the directrix equal to the distance to the focus and square:
    $\displaystyle (x+1)^2=(x+3)^2+(y-1)^2$
    $\displaystyle x^2+2x+1=x^2+6x+9+(y-1)^2$
    $\displaystyle -4x=(y-1)^2+8$
    $\displaystyle x=-\frac{1}{4}(y-1)^2-2$
    Standard form is a matter of opinion. You must follow the your notes/textbooks/instructor.
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