# Thread: Give equation in standard form of a parabola

1. ## 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.

2. ## Re: Give equation in standard form of a parabola

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3. ## Re: Give equation in standard form of a parabola

Originally Posted by Nfalconer
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.

4. ## 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

5. ## Re: Give equation in standard form of a parabola

Standard form x=-4(y-1) squared -2. Not sure if it is correct

6. ## Re: Give equation in standard form of a parabola

Originally Posted by Nfalconer
Standard form x=-4(y-1) squared -2. Not sure if it is correct
Well I get $\large{(y-1)^2=-4(x-2)}$

7. ## 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:

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

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

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

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

8. ## Re: Give equation in standard form of a parabola

Originally Posted by Plato
Well I get $\large{(y-1)^2=-4(x-2)}$
Originally Posted by MarkFL
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.