# Math Help - How to find coordinates of the focus and the equation of the directrix of the parabol

1. ## How to find coordinates of the focus and the equation of the directrix of the parabol

Hi I have an exercise and I’ve been struggling for a while with that. This is it:

Find coordinates of the focus and the equation of the directrix of the parabola whose equations is:

3y"squer"=8x
The chord which passes through the focus parallel to the directix is called the latus rectum of the parabola. Shown that the latus rectum of the above parabola has length8/3.

I do not understand this exercise at all. I have always thoughts that parabola equation looks like that:
y = 4x"squer" -4x +3
Anyone know how to resolve this problem?

2. ## Re: How to find coordinates of the focus and the equation of the directrix of the par

Originally Posted by leketapis
Hi I have an exercise and I’ve been struggling for a while with that. This is it:

Find coordinates of the focus and the equation of the directrix of the parabola whose equations is:

3y"squer"=8x
The chord which passes through the focus parallel to the directix is called the latus rectum of the parabola. Shown that the latus rectum of the above parabola has length8/3.

I do not understand this exercise at all. I have always thoughts that parabola equation looks like that:
y = 4x"squer" -4x +3
Anyone know how to resolve this problem?
$3y^2 = 8x$

$y^2 = \frac{8x}{3}$

this is a parabola with the x-axis as the axis of symmetry. the parabola opens to the right.

your text should have this form for a parabolic equation ...

$(y-k)^2 = 4p(x-h)^2$ , where $(h,k)$ are the coordinates of the vertex, $(h+p,k)$ are the coordinates of the focus, and $x = h-p$ is the directrix.

$(y-0)^2 = \frac{8}{3}(x - 0)^2$

vertex is at $(0,0)$

$4p = \frac{8}{3}$

$p = \frac{2}{3}$

$\left(\frac{2}{3} , 0\right)$ is the focus

$x = -\frac{2}{3}$ is the directrix

here is a link for you to investigate further ...

Conics: Parabolas: Introduction