parabola

Printable View

July 23rd 2006, 04:05 PM
mattballer082

this 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
July 23rd 2006, 05:18 PM
ThePerfectHacker

Quote:

Originally Posted by mattballer082

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

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

Let $(x,y)$ be a point on the parabola. Then it is equidistant from the focus $(0,2)$ and the line, $x=2$ . The distance to the focus by the distance formula is,
$s=\sqrt{x^2+(y-2)^2}$
Now, what is distance to the line $x=2$ . 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 $x=2$ is a vertical line, that means if you draw (and try to imagine this) a perpendicular line from the point $(x,y)$ you intersect at $(2,y)$ . Now the distance between these two points is,
$s=\sqrt{(2-x)^2+(y-y)^2}=\sqrt{(2-x)^2}$
By, definition these two distances are equal hence,
$\sqrt{x^2+(y-2)^2}=\sqrt{(2-x)^2}$
Square both sides,
$x^2+(y-2)^2=(2-x)^2$
Thus,
$x^2+y^2-4y+4=4-4x+x^2$
Kill the unneccesarry,
$y^2-4y=-4x$
Rewrite as,
$y^2+4x-4y=0$
July 24th 2006, 06:32 AM
Soroban

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

where $(h,k)$ is the Vertex of the parabola
and $p$ 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 $(1,2)\quad\Rightarrow\quad (h,k) = (1,2)$
. . the parabola opens to the left: $(y - k)^2 = 4p(x - h)$
. . distance from Vertex to Focus is $1$ to the left: $p = -1$

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