again, find the equation of the parabola with the given fetures.

focus (0,2) directrix x=2

please explain step by step if possible!

thank you

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- Jul 23rd 2006, 04:05 PMmattballer082this one is confusing me
again, find the equation of the parabola with the given fetures.

focus (0,2) directrix x=2

please explain step by step if possible!

thank you - Jul 23rd 2006, 05:18 PMThePerfectHackerQuote:

Originally Posted by**mattballer082**

**Definitnion:***A parabola is a set of point that are equidistant from a given point (focus) and a given line (directrix)*

Let $\displaystyle (x,y)$ be a point on the parabola. Then it is equidistant from the focus $\displaystyle (0,2)$ and the line, $\displaystyle x=2$. The distance to the focus by the distance formula is,

$\displaystyle s=\sqrt{x^2+(y-2)^2}$

Now, what is distance to the line $\displaystyle x=2$. This is slightly difficult. To find the distance from a point to a line you need to draw a perpendicular line, then find the distance of that line. Note, the $\displaystyle x=2$ is a vertical line, that means if you draw (and try to imagine this) a perpendicular line from the point $\displaystyle (x,y)$ you intersect at $\displaystyle (2,y)$. Now the distance between these two points is,

$\displaystyle s=\sqrt{(2-x)^2+(y-y)^2}=\sqrt{(2-x)^2}$

By, definition these two distances are equal hence,

$\displaystyle \sqrt{x^2+(y-2)^2}=\sqrt{(2-x)^2}$

Square both sides,

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

Thus,

$\displaystyle x^2+y^2-4y+4=4-4x+x^2$

Kill the unneccesarry,

$\displaystyle y^2-4y=-4x$

Rewrite as,

$\displaystyle y^2+4x-4y=0$ - Jul 24th 2006, 06:32 AMSoroban
Hello, mattballer0821

Quote:

Find the equation of the parabola with the given features.

Focus (0,2) directrix x=2

You're expected to know these basic facts . . .

The Vertex is halfway between the Focus and the Directrix.

There are two forms:

. . . . Vertical: $\displaystyle (x-h)^2 = 4p(y-k)$ . . . opens up or down: $\displaystyle \cup$ or $\displaystyle \cap$

. . Horizontal: $\displaystyle (y-k)^2 = 4p(x-h)$ . . . opens right or left: $\displaystyle \subset$ or $\displaystyle \supset$

where $\displaystyle (h,k)$ is the Vertex of the parabola

and $\displaystyle p$ is the*directed*distance from the Vertex to the Focus.

The parabola always "bends around" the Focus.

The parabola always "bends away" from the Directrix.

Sketch the given information:Code:`o |`

|o :

| o :

F(0,2)* oV :

| o :

- - - - +o- - - + -

o | :

| -x=2

We see that the Vertex is $\displaystyle (1,2)\quad\Rightarrow\quad (h,k) = (1,2)$

. . the parabola opens to the left: $\displaystyle (y - k)^2 = 4p(x - h)$

. . distance from Vertex to Focus is $\displaystyle 1$ to the__left__: $\displaystyle p = -1$

Therefore: .$\displaystyle (y - 2)^2 \,= \,-4(x - 1)$