1. ## i dont get it!! ahhh! help

find the equation of the parabola with the given features. graph the parabola.

1.Vertex (0,0) focus (0,3)

and

2. (0,0) directrix x+1=0

Both of them confuse me. can someone explain it step by step please!!!

2. Originally Posted by mattballer082
find the equation of the parabola with the given features. graph the parabola.

1.Vertex (0,0) focus (0,3)

and

2. (0,0) directrix x+1=0

Both of them confuse me. can someone explain it step by step please!!!
Wikipedia tells us that a parabola with vertical symetry axis (which is what you
have in your first problem) with vertex $(h,k)$, and focus $(h,k+p)$ has directrix
$y=k-p$, and equation:

$
(x-h)^2=4p(y-k)
$

In your first question the vertex is at $(0,0)$ so $h=0, \ k=0$, and as the focus
is at $(0,3)$ so $p=3$, and the equation becomes:

$
x^2=12y
$

In your second problem the symetry axis is horizontal, so what I will do
is derive the equation as though it were vertical and finaly interchenge $x$ and $y$.

Flipping the axis leaves the vertex at $(0,0)$ but the directrix becomes $y+1=0$
or equivalently $y=-1$

In this case $h=0, \ k=0$ and $p=1$ so the equation
for the vertical axis parabola is:

$
x^2=4y
$
,

and interchanging $x$ and $y$ gives us:

$
y^2=4x
$

for the horizontal axis parabola.

RonL

3. Hello, mattballer082!

These problems have the vertex at the origin.

There are two types:
. . $x^2 = 4py$ . . . the parabola opens up or down: $\cup$ or $\cap$
. . $y^2 = 4px$ . . . the parabola opens right or left: $\subset$ or $\supset$

$p$ is the distance from the Vertex to the Focus
. . and the distance from the Vertex to the Directrix.

Note: the curve always bends "around the Focus" and "away from the Directrix".

Find the equation of the parabola with the given features.
Graph the parabola.

1. Vertex (0,0), focus (0,3)

Make a sketch.
Code:
            |
*     Fo(0,3) *
*     |     *
*   |   *
- - - o - - -
V|(0,0)

Plot the Vertex and the Focus.
We know that the parabola bends around the Focus,
. . so we know its orientation: opens upward.

We use the form: . $x^2 = 4py$

We see that $p = 3$
. . Therefore, the equation is: . $x^2 = 12y$

2. Vertex (0,0), directrix x + 1 = 0

The vertex is at the origin; the directrix is the vertical line: $x = -1$
Code:
          :   |         *
:   |   *
:   |*
- - + - o - - - - - - -
:   |*
:   |   *
:   |         *
x=-1

Plot the Vertex and the Directrix.
We the know the parabola "bends away from the directrix"
. . so we know its orientation: opens right.

We use the form: . $y^2 = 4px$

We see that $p = 1$
. . Therefore, the equation is: . $y^2 = 4x$