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!!!

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- Jul 20th 2006, 08:49 PMmattballer082i 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!!! - Jul 20th 2006, 10:33 PMCaptainBlackQuote:

Originally Posted by**mattballer082**

have in your first problem) with vertex $\displaystyle (h,k)$, and focus $\displaystyle (h,k+p)$ has directrix

$\displaystyle y=k-p$, and equation:

$\displaystyle

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

$

In your first question the vertex is at $\displaystyle (0,0)$ so $\displaystyle h=0, \ k=0$, and as the focus

is at $\displaystyle (0,3)$ so $\displaystyle p=3$, and the equation becomes:

$\displaystyle

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 $\displaystyle x$ and $\displaystyle y$.

Flipping the axis leaves the vertex at $\displaystyle (0,0)$ but the directrix becomes $\displaystyle y+1=0$

or equivalently $\displaystyle y=-1$

In this case $\displaystyle h=0, \ k=0$ and $\displaystyle p=1$ so the equation

for the vertical axis parabola is:

$\displaystyle

x^2=4y

$,

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

$\displaystyle

y^2=4x

$

for the horizontal axis parabola.

RonL - Jul 21st 2006, 06:45 AMSoroban
Hello, mattballer082!

These problems have the vertex at the origin.

There are two types:

. . $\displaystyle x^2 = 4py$ . . . the parabola opens up or down: $\displaystyle \cup$ or $\displaystyle \cap$

. . $\displaystyle y^2 = 4px$ . . . the parabola opens right or left: $\displaystyle \subset$ or $\displaystyle \supset$

$\displaystyle p$ 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".

Quote:

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: .$\displaystyle x^2 = 4py$

We see that $\displaystyle p = 3$

. . Therefore, the equation is: .$\displaystyle x^2 = 12y$

Quote:

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

The vertex is at the origin; the directrix is the vertical line: $\displaystyle x = -1$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: .$\displaystyle y^2 = 4px$

We see that $\displaystyle p = 1$

. . Therefore, the equation is: .$\displaystyle y^2 = 4x$